Bài 1:
a)5x(x-2)+(2-x)=0
b)x(2x-5)-5x+10
c)(x3-x2)-4x2+8x-4=0
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9 tháng 10 2021
\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)
\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)
\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
17 tháng 10 2021
\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)
\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)
\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)
T
3 tháng 10 2023
Bài 4.
\(A=2x^3+(x+1)^3-3x(x-2)(x+2)-3(x^2+5x+9)\\=2x^3+(x^3+3x^2+3x+1)-3x(x^2-4)-3x^2-15x-27\\=2x^3+x^3+3x^2+3x+1-3x^3+12x-3x^2-15x-27\\=(2x^3+x^3-3x^3)+(3x^2-3x^2)+(3x+12x-15x)+(1-27)\\=-26\\---\)
\(B=x(x-4x)+x(2-x)(x+2)+4(2x^2-5x+4)\\=x\cdot(-3x)+x(2-x)(2+x)+8x^2-20x+16\\=-3x^2+x(4-x^2)+8x^2-20x+16\\=-3x^2+4x-x^3+8x^2-20x+16\)
Bạn kiểm tra lại đề giúp mình!
\(C=(x-2y)(x^2+2xy+4y^2)-(x^3-8y^3+10)\) (sửa đề)
\(=x^3-(2y)^3-x^3+8y^2-10\\=x^3-8y^3-x^3+8y^3-10\\=(x^3-x^3)+(-8y^3+8y^3)-10\\=-10\)
Bài 5.
\(d)xy^2-3x^3y^2-2x(xy-3xy^2)\\=xy^2-3x^3y^2-2x^2y+6x^2y^2\\---\\f)(x-y)(2x+y)-2x^2+y^2+3xy\\=x(2x+y)-y(2x+y)-2x^2+y^2+3xy\\=2x^2+xy-2xy-y^2-2x^2+y^2+3xy\\=(2x^2-2x^2)+(xy-2xy+3xy)+(-y^2+y^2)\\=2xy\)
\(Toru\)
KV
2 tháng 10 2020
a) (2x2 - x) + 4x - 2 = 0
x(2x - 1) + 2(2x - 1) = 0
(2x - 1)(x + 2) = 0
2x - 1 = 0 hoặc x + 2 = 0
* 2x - 1 = 0
2x = 1
x = \(\frac{1}{2}\)
* x + 2 = 0
x = -2
Vậy x = -2; x = \(\frac{1}{2}\)
b) x2 - 6x + 8 = 0
x2 - 2x - 4x + 8 = 0
(x2 - 2x) + (-4x + 8) = 0
x(x - 2) - 4(x - 2) = 0
(x - 2)(x - 4) = 0
x - 2 = 0 hoặc x - 4 = 0
* x - 2 = 0
x = 2
* x - 4 = 0
x = 4
Vậy x = 2; x = 4
c) x4 - 8x2 - 9 = 0
x4 + x2 - 9x2 - 9 = 0
(x4 - 9x2) + (x2 - 9) = 0
x2(x2 - 9) + (x2 - 9) = 0
(x2 - 9)(x2 + 1) = 0
x2 - 9 = 0 (vì x2 + 1 > 0 với mọi x)
x2 = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
DT
13
VN
Võ Ngọc Phương
CTVHS
VIP
26 tháng 8 2023
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
26 tháng 8 2023
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
a, 5x(x-2) + (2-x)=0
⇔5x(x-2) - (x-2) =0
⇔(x-2)(5x-1)=0
\(\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy....
c, (x3 - x2) - 4x2 + 8x -4 =0
⇔x3 - x2 -4x2 + 8x - 4=0
⇔x2(x-1) - 4x(x-1) +4(x-1) =0
⇔(x-1) (x-2)2=0
⇔\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
Phần b cậu có chép sai đề không?
.chỗ đó là giải phương trình hay PTĐTTNT vậy?