a) 2 |x-1| 4 b) |2x-3|-x=5

Nguyễn Văn Quân

7 tháng 5 2022

b) +) |2x-3|=2x-3 khi 2x-3 ≥ 0 ⇔ x ≥ $\dfrac{3}{2}$
Khi đó PT b trở thành :
2x-3-x=5
⇔2x-x=5+3
⇔x=8 (tm)
+) |2x-3|=-2x+3 khi 2x-3 < 0 ⇔ x < $\dfrac{3}{2}$
Khi đó PT b trở thành :
-2x+3-x=5
⇔-2x-x=5-3
⇔-3x=2
⇔x=$-\dfrac{2}{3}$ (tm)
vậy tập nghiệm của PT là S={8;$-\dfrac{2}{3}$ }

Nguyễn Danh Hoàng Minh

7 tháng 5 2022

a) kiểm tra xem có sai đề không bạn nhé

b) |2x-3|-x=5

|2x-3| =5+x

TH1: 2x-3 > 0 => x>3/2

=> 2x-3=5+x

=> x= 8 <t/m>

TH2 : 2x-3<0 => x<3/2

=> -2x+3=5+x

=> x =$-\dfrac{3}{2}$

Câu 1 : Giải phương trình a. 5(x-3)-4=2(x-1) b. 5-(6-x)=4(3-2x) c. (3x+5)(2x+1)=(6x-2)(x-3) d. (x+2)2 + 2(x-4)=(x-4)(x-2) Bài 2 : Giải phương trình a) x/3 - 5x/6 - 15x/12 = x/4 - 5 b) 8x-3/4 - 3x-2/2 = 2x-1/2 + x+3/4 c) x-1/2 - x+1/15 - 2x-13/6 = 0 d) 3(3-x)/8 + 2(5-x)/3 = 1-x/2 - 2 e) 3(5x-2)/4 - 2 = 7x/3 - 5(x-7) Bài 3 Giải phương trình a) (5x-4)(4x+6)=0 b) (x-5)(3-2x)(3x+4)=0 c) (2x+1)(x2+2)=0 d) (8x-4)(x2+2x+2)=0 Bài 4 Giải phương trình a)...

Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!

Trường Beenlee

18 tháng 2 2020

Đọc tiếp

Nguyễn Lê Phước Thịnh

18 tháng 2 2020

Bài 1:

a) 5(x-3)-4=2(x-1)

$\Leftrightarrow5x-15-4=2x-2$

$\Leftrightarrow5x-19-2x+2=0$

$\Leftrightarrow3x-17=0$

$\Leftrightarrow3x=17$

$\Leftrightarrow x=\frac{17}{3}$

Vậy: $x=\frac{17}{3}$

b) 5-(6-x)=4(3-2x)

$\Leftrightarrow5-6+x=12-8x$

$\Leftrightarrow-1+x-12+8x=0$

$\Leftrightarrow-13+9x=0$

$\Leftrightarrow9x=13$

$\Leftrightarrow x=\frac{13}{9}$

Vậy: $x=\frac{13}{9}$

c) (3x+5)(2x+1)=(6x-2)(x-3)

$\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6$

$\Leftrightarrow6x^2+13x+5=6x^2-20x+6$

$\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0$

$\Leftrightarrow33x-1=0$

$\Leftrightarrow33x=1$

$\Leftrightarrow x=\frac{1}{33}$

Vậy: $x=\frac{1}{33}$

d) $\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)$

$\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8$

$\Leftrightarrow x^2+6x-4=x^2-6x+8$

$\Leftrightarrow x^2+6x-4-x^2+6x-8=0$

$\Leftrightarrow12x-12=0$

$\Leftrightarrow x=1$

Vậy:x=1

Bài 2:

a)$\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5$

$\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0$

$\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0$

$\Leftrightarrow4x-10x-15x-3x+60=0$

$\Leftrightarrow-24x+60=0$

$\Leftrightarrow-24x=-60$

$\Leftrightarrow x=\frac{5}{2}$

Vậy: $x=\frac{5}{2}$

b) $\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}$

$\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0$

$\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0$

$\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0$

$\Leftrightarrow8x-3-6x+4-4x+2-x-3=0$

$\Leftrightarrow-3x=0$

$\Leftrightarrow x=0$

Vậy: x=0

c) $\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0$

$\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0$

$\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0$

$\Leftrightarrow15x-15-2x-2-10x+65=0$

$\Leftrightarrow3x+48=0$

$\Leftrightarrow3x=-48$

$\Leftrightarrow x=-16$

Vậy: x=-16

d) $\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2$

$\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0$

$\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0$

$\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0$

$\Leftrightarrow27-9x+80-16x-12+12x+48=0$

$\Leftrightarrow-13x+143=0$

$\Leftrightarrow-13x=-143$

$\Leftrightarrow x=11$

Vậy: x=11

e) $\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)$

$\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0$

$\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0$

$\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0$

$\Leftrightarrow45x-18-24-28x+60x-420=0$

$\Leftrightarrow77x-462=0$

$\Leftrightarrow77x=462$

$\Leftrightarrow x=6$

Vậy:x=6

Bài 3:

a) $\left(5x-4\right)\left(4x+6\right)=0$

$\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0$

Vì $2\ne0$

nên $\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.$

Vậy: $x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}$

b) $\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.$

Vậy: $x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}$

c) $\left(2x+1\right)\left(x^2+2\right)=0$

Ta có: $\left(2x+1\right)\left(x^2+2\right)=0$(1)

Ta có: $x^2\ge0\forall x$

$\Rightarrow x^2+2\ge2\ne0\forall x$(2)

Từ (1) và (2) suy ra:

$2x+1=0$

$\Leftrightarrow2x=-1$

$\Leftrightarrow x=\frac{-1}{2}$

Vậy: $x=\frac{-1}{2}$

d) $\left(8x-4\right)\left(x^2+2x+2\right)=0$

$\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0$

Ta có: $x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1$

Ta lại có $\left(x+1\right)^2\ge0\forall x$

$\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x$(3)

Ta có: $4\ne0$(4)

Từ (3) và (4) suy ra

2x-1=0

$\Leftrightarrow2x=1$

$\Leftrightarrow x=\frac{1}{2}$

Vậy: $x=\frac{1}{2}$

Bài 4:

a) $\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)$

$\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2$

$\Leftrightarrow2x^2-x-6=x^2-3x+2$

$\Leftrightarrow2x^2-x-6-x^2+3x-2=0$

$\Leftrightarrow x^2+2x-8=0$

$\Leftrightarrow x^2+2x+1-9=0$

$\Leftrightarrow\left(x+1\right)^2-3^2=0$

$\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x+4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.$

Vậy: $x\in\left\{2;-4\right\}$

b) $\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)$

$\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0$

$\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0$

$\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0$

$\Leftrightarrow\left(x-4\right)\cdot3x=0$

Vì $3\ne0$

nên $\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.$

Vậy: $x\in\left\{0;4\right\}$

c) $9x^2-1=\left(3x+1\right)\left(2x-3\right)$

$\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0$

$\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0$

$\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0$

$\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.$

Vậy: $x\in\left\{-\frac{1}{3};-2\right\}$

d) $\left(x+2\right)^2=9\left(x^2-4x+4\right)$

$\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0$

$\Leftrightarrow x^2+4x+4-9x^2+36x-36=0$

$\Leftrightarrow-8x^2+40x-32=0$

$\Leftrightarrow-\left(8x^2-40x+32\right)=0$

$\Leftrightarrow-8\left(x^2-5x+4\right)=0$

Vì $-8\ne0$

nên $x^2-5x+4=0$

$\Leftrightarrow x^2-x-4x+4=0$

$\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0$

$\Leftrightarrow\left(x-1\right)\left(x-4\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.$

Vậy: $x\in\left\{1;4\right\}$

e) $4\left(2x+7\right)^2-9\left(x+3\right)^2=0$

$\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0$

$\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0$

$\Leftrightarrow7x^2+58x+115=0$

$\Leftrightarrow7x^2+23x+35x+115=0$

$\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0$

$\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.$

Vậy: $x\in\left\{\frac{-23}{7};-5\right\}$

Bài 5:

a) $\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)$

$\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0$

$\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0$

$\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0$

$\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0$

$\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.$

Vậy: $x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}$

b) $\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)$

$\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x$

$\Leftrightarrow2x^2-2x=-x^2-2x+3$

$\Leftrightarrow2x^2-2x+x^2+2x-3=0$

$\Leftrightarrow3x^2-3=0$

$\Leftrightarrow3\left(x^2-1\right)=0$

$\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0$

Vì $3\ne0$

nên $\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.$

Vậy: $x\in\left\{1;-1\right\}$

c) $x^4+x^3+x+1=0$

$\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=0$

$\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0$(5)

Ta có: $x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}$

Ta lại có: $\left(x-\frac{1}{2}\right)^2\ge0\forall x$

$\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x$(6)

Từ (5) và (6) suy ra

$\left(x+1\right)^2=0$

$\Leftrightarrow x+1=0$

$\Leftrightarrow x=-1$

Vậy: x=-1

Trần Quốc Khanh

18 tháng 2 2020

ko khó đâu, chủ yếu nhát làm

Nguyệt Hi

21 tháng 6 2021

Tìm x: a)(2x+1)(1-2x)+(1-2x)^2=18

b) 2(x+1)^2-(x-3)(x+3)-(x-4)^2=0

c) (x-5)^2-x(x-4)=9

d) (x-5)^2+(x-4)(1-x)=0

Nguyễn Nho Bảo Trí

21 tháng 6 2021

a) (2x + 1)(1 - 2x) + (1 - 2x)² = 18

= ( 1 - 2x) $\left[\left(2x+1+1-2x\right)\right]$ = 18

= 2(1 - 2x) - 18 = 0

= 2 - 4x - 18 = 0

= -16 - 4x = 0

= -4x = 16

= x = $\dfrac{16}{-4}=-4$

b) 2(x + 1)² -(x - 3)(x + 3) - (x - 4)² = 0

= 2 (x² + 2x + 1) - (x² - 9) - (x² - 8x + 16) = 0

= 2x² + 4x + 2 - x² + 9 - x² + 8x - 16 = 0

= 12x - 5 = 0

= 12x = 5

= x = $\dfrac{5}{12}$

c) (x - 5)² - x(x - 4) = 9

= x² - 10x + 25 - x² + 4x - 9 = 0

= -6x + 16 = 0

= -6x = -16

= x = $\dfrac{-16}{-6}=\dfrac{8}{3}$

d) (x - 5)² + (x - 4)(1 - x)

= x² - 10x + 25 + 5x - x² - 4 = 0

= -5x + 21 = 0

= -5x = -21

= x = $\dfrac{-21}{-5}=\dfrac{21}{5}$

Chúc bạn học tốt

Đúng(2)

Lê Đức Hoàng

16 tháng 3 2023

giúp mik vs mai mik kiểm tra rùi
a) $\frac{x-1}{x}$ - $\frac{1}{x+1}$ = $\frac{2x-1}{x2+x}$
b) (x+2).(5-3x)=0
c)$\frac{5(1-2x)}{3}$ + $\frac{x}{2}$ = $\frac{3(x-5)}{4}$ - 2
d)$(x+2)^{2}$ - (x-1).(x+3) = (2x-4).(x+4)-3
e)$(2x-3)^{2}$ = (2x-3).(x+1)

Nguyễn Lê Phước Thịnh

16 tháng 3 2023

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>$x=\dfrac{-3\pm\sqrt{61}}{2}$

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

nguyễn vương hải

3 tháng 10 2021

phân tích đa thức thành nhân tử

a) 4x (a-b) +6xy(b-a)

b) (6x+3) - ( 2x-5) (2x+1)

c) 4 ( x-3)^2 +2x (3-x)

d) x^4 +2x^2 -4x-4

e) 2x (x+y) -x -y

g)( 3x-1 )^2 - (x+3)^2

Minh Hiếu

3 tháng 10 2021

a) $4x\left(a-b\right)+6xy\left(b-a\right)$

$=4x\left(a-b\right)-6xy\left(a-b\right)$

$=\left(4x-6xy\right)\left(a-b\right)$

$=2x\left(2-3y\right)\left(a-b\right)$

Minh Hiếu

3 tháng 10 2021

b) $\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)$

$=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)$

$=\left(3-2x+5\right)\left(2x+1\right)$

$=\left(8-2x\right)\left(2x+1\right)$

$=2\left(4-x\right)\left(2x+1\right)$

Vy Pé

24 tháng 2 2020

a.(x+5)(2x-1)=(2x-3)(x+1)

b.(x+1)(x+9)=(x+3)(x+5)

c.(3x+5)(2x+1)=(6x-2)(x-3)

d.(x-2)(3x+5)=(2x-4)(x+1)

đ.9x²-1=(3x+1)(2x-3)

e.(2x+5)(x-4)=(x-5)(4-x)

Nguyễn Lê Phước Thịnh

24 tháng 2 2020

a) Ta có: $\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)$

$\Leftrightarrow\left(x+5\right)\left(2x-1\right)-\left(2x-3\right)\left(x+1\right)=0$

$\Leftrightarrow2x^2-x+10x-5-\left(2x^2+2x-3x-3\right)=0$

$\Leftrightarrow2x^2+9x-5-2x^2+x+3=0$

$\Leftrightarrow10x-2=0$

hay 10x=2

$\Leftrightarrow x=\frac{1}{5}$

Vậy: $x=\frac{1}{5}$

b) Ta có: $\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)$

$\Leftrightarrow x^2+9x+x+9=x^2+5x+3x+15$

$\Leftrightarrow x^2+10x+9-x^2-8x-15=0$

$\Leftrightarrow2x-6=0$

hay 2x=6

$\Leftrightarrow x=3$

Vậy: x=3

c) Ta có: $\left(3x+5\right)\left(2x+1\right)=\left(6x-2\right)\left(x-3\right)$

$\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6$

$\Leftrightarrow6x^2+13x+5=6x^2-20x+6$

$\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0$

$\Leftrightarrow33x-1=0$

$\Leftrightarrow33x=1$

hay $x=\frac{1}{33}$

Vậy: $x=\frac{1}{33}$

d) Ta có: $\left(x-2\right)\left(3x+5\right)=\left(2x-4\right)\left(x+1\right)$

$\Leftrightarrow3x^2+5x-6x-10=2x^2+2x-4x-4$

$\Leftrightarrow3x^2-x-10=2x^2-2x-4$

$\Leftrightarrow3x^2-x-10-2x^2+2x+4=0$

$\Leftrightarrow x^2+x-6=0$

$\Leftrightarrow x^2+3x-2x-6=0$

$\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0$

$\Leftrightarrow\left(x+3\right)\left(x-2\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.$

Vậy: $x\in\left\{-3;2\right\}$

đ) Ta có: $9x^2-1=\left(3x+1\right)\left(2x-3\right)$

$\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0$

$\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0$

$\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0$

$\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0$

Vậy: $x\in\left\{-\frac{1}{3};-2\right\}$

e) Ta có: $\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)$

$\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0$

$\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0$

$\Leftrightarrow\left(x-4\right)\cdot3x=0$

Vì $3\ne0$

nên $\left[{}\begin{matrix}x-4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.$

Vậy: $x\in\left\{0;4\right\}$

Akai Haruma

Giáo viên

24 tháng 2 2020

a) $(x+5)(2x-1)=(2x-3)(x+1)$

$\Leftrightarrow 2x^2+9x-5=2x^2-x-3$

$\Leftrightarrow 10x=2\Rightarrow x=\frac{1}{5}$

$(x+1)(x+9)=(x+3)(x+5)$

$\Leftrightarrow x^2+10x+9=x^2+8x+15$

$\Leftrightarrow 2x=6\Rightarrow x=3$

$(3x+5)(2x+1)=(6x-2)(x-3)$

$\Leftrightarrow 6x^2+13x+5=6x^2-20x+6$

$\Leftrightarrow 33x=1\Rightarrow x=\frac{1}{33}$

giải pt: a)2x+1/6 - x-2/4= 3-2x/3 -x b)2x/x-1 + 4/x^2+2x-3 = 2x-5/x+3

Ngô Dương

29 tháng 3 2018

Aki Tsuki

29 tháng 3 2018

Violympic toán 8

thuc quyen thái

8 tháng 12 2021

làm phép chia :

a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)

b) (x^3 -8) : (x^2 + 2x +4)

c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2

d) (-2x^5 + 3x^2 - 4x^3) :2x^2

e) (15x^3 - 10x^2 + x - 2) : (x - 2)

f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)

Nguyễn Lê Phước Thịnh

8 tháng 12 2021

b: =x-2

d: $=-x^3+\dfrac{3}{2}-2x$

Huỳnh Xương Hưng

14 tháng 12 2021

a) x+1/2x-6-4/2x-6 b) 3x-4/6x+3-x-5/6x+3

c) x-1/x-3-3x-8/3-x+3-2x/x-3 d) 3/x+5-5/x-7

e) 3/x+5-5/x-7 f) 2/x-2+3/x+2+5x-18/x²-4

Trương Thị Trang Thư

10 tháng 8 2020

Rút gọn biểu thức sau:

a. A= ( 5- x ). (x + 5) - 2.( x - 1). ( x - 3) - 3.( x - 2)²

b. B= ( 3 - 2x).( x - 2) + ( 2x - 5)² - ( x - 4)

c. C= ( x - 4). ( x - 2) - 3.( x - 2) . ( 3 - 2x) - ( 2x + 1)²

d D= 2.( x - 1)² - 3.( x - 1). ( x + 2) - ( 2x + 1)²

Phú Hưng (Phú và Hưng)

10 tháng 8 2020

mấy bài này ko khó, mong bạn tự làm được :)

Trúc Giang

10 tháng 8 2020

a) $A=\left(5-x\right)\left(x+5\right)-2\left(x-1\right)\left(x-3\right)-3\left(x-2\right)^2$

$=\left(5-x\right)\left(5+x\right)-\left(2x-2\right)\left(x-3\right)-3\left(x^2-2.2x+2^2\right)$

$=\left(5^2-x^2\right)-\left[2x\left(x-3\right)-2\left(x-3\right)\right]-3\left(x^2-4x+4\right)$

$=25-x^2-\left[\left(2x^2-6x\right)-\left(2x-6\right)\right]-3x^2+12x-12$

$=25-x^2-\left(2x^2-6x-2x+6\right)-3x^2+12x-12$

$=25-x^2-2x^2+6x+2x-6-3x^2+12x-12$

$=7+20x-6x^2$

b/ $B=\left(3-2x\right)\left(x-2\right)+\left(2x-5\right)^2-\left(x-4\right)$

$=3\left(x-2\right)-2x\left(x-2\right)+\left[\left(2x\right)^2-2.2x.5+5^2\right]-x+4$

$=3x-6-2x^2+4x+4x^2-20x+25-x+4$

$=23-14x+2x^2$

c/ $C=\left(x-4\right)\left(x-2\right)-3\left(x-2\right)\left(3-2x\right)-\left(2x+1\right)^2$

$=x\left(x-2\right)-4\left(x-2\right)-\left(3x-6\right)\left(3-2x\right)-\left[\left(2x\right)^2+2.2x.1+1^2\right]$

$=x^2-2x-4x+8-\left[3x\left(3-2x\right)-6\left(3-2x\right)\right]-4x^2-4x-1$

$=x^2-2x-4x+8-\left(9x-6x^2-18+12x\right)-4x^2-4x-1$

$=x^2-2x-4x+8-9x+6x^2+18-12x-4x^2-4x-1$

$=25-31x+3x^2$

d/ $D=2\left(x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(2x+1\right)^2$

$=2.\left(x^2-2x+1\right)-\left(3x-3\right)\left(x+2\right)-\left[\left(2x\right)^2+2.2x+1\right]$

$=2x^2-4x+2-\left[3x\left(x+2\right)-3\left(x+2\right)\right]-\left(4x^2+4x+1\right)$

$=2x^2-4x+2-\left(3x^2+6x-3x-6\right)-\left(4x^2+4x+1\right)$

$=2x^2-4x+2-3x^2-6x+3x+6-4x^2-4x-1$

$=7-11x-5x^2$

P/s: Ko chắc ạ!

Thân Bích Ngọc

12 tháng 8 2019

1, tìm x :

a) (x+2).(x+3)-(x-2).(x+5)=0

b) (2x+3).(x-4)+(x-5).(x-2)=(x-4).(3x-5)

c) (3x-5).(7-5x)-(5x+2).(2-3x)=3

d)(x-7).(x+7)-(x-4)2-x=10+3

e) (x+1/2)2-(x-1).(x+1)=2x-1

2, Thu gọn :

a) (x+5)2-(3-x)2-2(x-4).(x+4)

b)4(x-1)2+(2x+3)2-8(x-2)(x+2)

lê thị hương giang

12 tháng 8 2019

Trần Thanh Phương

12 tháng 8 2019

lê thị hương giang cho hỏi bạn học lớp mấy ?