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15 tháng 11 2021

a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)

15 tháng 11 2021

a/ \(x^2-25=0\)

\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)

 

b/ \(x\left(x+7\right)+x+7=0\)

\(x\left(x+7\right)+\left(x+7\right)=0\)

\(\left(x+7\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

22 tháng 12 2021

a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

21 tháng 7 2021

Bài 10:

a) (x+2)2 -x(x+3) + 5x = -20

=> x2 + 4x + 4 - x2 - 3x + 5x = -20

=> 6x = -20 + (-4)

=> 6x = -24

=> x = -4

b) 5x3-10x2+5x=0   

=>5x(x2-2x+1)=0

=>5x(x-1)2 =0

=> 5x=0 hoặc (x-1)2=0

=>x=0 hoặc x=1

c) (x- 1)- (x+ x+ 1)(x- 1) = 0

=> (x2 - 1)[(x- 1)2 -  (x+ x+ 1)] = 0

<=> (x2 - 1)(x4 - 2x2 + 1 - x- x- 1) = 0

<=>  (x2 - 1)(-3x2) = 0

<=> (x2 - 1)=0 hoặc (-3x2) =0

<=> x2=1 hoặc x2=0

<=> x=−1;1 hoặc x=0

d)

(x+1)3−(x−1)3−6(x−1)2=-19

⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0

⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0

⇔12x+13=0⇔12x+13=0

⇔12x=-13

⇔x=-23/12

Học tốt nhé:333banhqua

 

 

 

18 tháng 8 2021

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

a: 49x^2-25=0

=>(7x-5)(7x+5)=0

=>7x-5=0 hoặc 7x+5=0

=>x=5/7 hoặc x=-5/7

b: Đề thiếu vế phải rồi bạn

c: (3x-2)^2-9(x+4)(x-4)=2

=>9x^2-12x+4-9(x^2-16)=2

=>9x^2-12x+4-9x^2+144=2

=>-12x+148=2

=>-12x=-146

=>x=146/12=73/6

d: x^3-6x^2+12x-8=0

=>(x-2)^3=0

=>x-2=0

=>x=2

e: x^3-9x^2+27x-27=0

=>(x-3)^3=0

=>x-3=0

=>x=3

3 tháng 9 2023

a) \(-25+49x^2=0\)

\(\Leftrightarrow49x^2-25=0\)

\(\Leftrightarrow\left(7x\right)^2-5^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)

b) \(16x^2-25\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)

\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)

c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)

\(\Leftrightarrow-84x-140=2\)

\(\Leftrightarrow-84x=142\)

\(\Leftrightarrow x=-\dfrac{142}{84}\)

\(\Leftrightarrow x=-\dfrac{71}{42}\)

d) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

e) \(-27+27x-9x^2+x^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

1 tháng 7 2021

a)4x2+4x+1-x2-10x-25=0

`<=>(2x+1)^2-(x+5)^2=0`

`<=>(2x+1-x-5)(2x+1+x+5)=0`

`<=>(x-4)(3x+6)=0`

`<=>(x-4)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\) 
b)(x^2+x+7)(x^2+x-7)=(x2+x)2-7x

`<=>(x^2+x)^2-7^2=(x^2+x)^2-7x`

`<=>-7^2=-7x`

`<=>-49=-7x`

`<=>x=7`

Vậy x=7

a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow x^2+7x-x^2-x+6=0\)

hay x=-1

b: Ta có: \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

4 tháng 9 2021

b. (x + 2)2 - x2 + 4 = 0

<=> (x + 2 - x)(x + 2 + x) + 4 = 0

<=> 2(2 + 2x) + 4 = 0

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

1 tháng 11 2021

a) \(\Rightarrow\left(2x-3\right)^2=49\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

1 tháng 11 2021

a, ⇒ (2x - 3)2 = 49

    ⇒  (2x - 3)2 = \(\left(\pm7\right)^2\)

    ⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0

    ⇒ (x - 5).(2x + 7)  = 0

    ⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c, ⇒ x2 - 5x + 2x - 10 = 0

    ⇒ (x2 - 5x) + (2x - 10) = 0

    ⇒ x.(x - 5) +2.(x - 5)    = 0

    ⇒ (x - 5).(x + 2)=0

    \(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)

22 tháng 10 2021

\(\left(2x-3\right)^2=7^2\)

\(2x-3=7\)

\(2x=10\)

\(x=5\)

Vậy x=5

22 tháng 10 2021

a: \(\left(2x-3\right)^2-49=0\)

\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)