Ai giup em voi a :)

đéo hiểu đề

Giúp em với ạ T_T

8) \(\left(x+4\right)\left(6x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\6x=12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy \(x\in\left\{-4;2\right\}\)

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}-0\\3x=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\x=-\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=-\frac{1}{9}\end{cases}}}\)

Vậy \(x\in\left\{\frac{7}{16};-\frac{1}{9}\right\}\)

12) \(3x-2x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)

13) \(5x+10x^2=0\)

\(\Leftrightarrow5x\left(1+2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x\in\left\{0;-\frac{1}{2}\right\}\)

8) (x+4)(6x-12)=0

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy x={-4;2}

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\3x=\frac{-1}{3}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=\frac{-1}{9}\end{cases}}}\)

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

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

\(x^3-27+4x-x^3=1\)

\(4x-27=1\)

\(4x=28\)

\(x=7\)

Vậy x = 7

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

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

\(\Rightarrow x^3-27+4x-x^3=1\)

\(\Rightarrow4x-27=1\)

\(\Rightarrow4x=28\)

\(\Rightarrow x=7\)

Vậy \(x=7\)

+)   (5x-1). (2x+3)-3. (3x-1)=0

10x^2+15x-2x-3 - 9x+3=0

10x^2 +8x=0

2x(5x+4)=0

=> x=0 hoặc x= -4/5

+)    x^3 (2x-3)-x^2 (4x^2-6x+2)=0

2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0

-2x^4 + 3x^3-2x^2=0

x^2(-2x^2+x-2)=0

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

=> x=0 hoặc x=1

+)   x (x-1)-x^2+2x=5

x^2 -x -x^2+2x=5

x=5

+)     8 (x-2)-2 (3x-4)=25

8x - 16-6x+8=25

2x=33

x=33/2

= x³ + 3³ -x(x² -1) -27 =0 ( tổng các lập phuong)

x =0 

CX100%

bạn chỉ cần phá hết hằng đẳng thức ra thôi 

câu 1

(x+3)^2+(x-2)(x+2)-2(x-1)^2=7

<=> x^2+6x+9+x^2-4-2x^2-4x-2-7=0

<=> x^2+x^2-2x^2+6x-4x+9-4-2-7=0

<=> 2x-4=0

<=> x=2

câu 2

(x+3)(x^2-3x+9)-x(x-1)(x+1)-27=0

<=> x^3+27-x(x^2-1)-27=0

<=> x^3-x^3+x+27-27=0

<=>x=0

OLM chỉ có phần chụp ảnh cho CTV

Lưu ý bạn cố phải viết thẳng hàng vì OLM ko viết đc

a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)

b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)

d,x² - 4x + 3 = x² - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

e,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

f,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2

\(\text{1) }\)

\(\text{a) }5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y+2z\right)\left(x-y-2z\right)\)

\(\text{b) }5x^2+5xy-x-y\)

\(=\left(5x^2-x\right)+\left(5xy-y\right)\)

\(=x\left(5x-1\right)+y\left(5x-1\right)\)

\(=\left(5x-1\right)\left(x+y\right)\)

\(\text{c) }2\left(x+4\right)-x^2+16\)

\(=2\left(x+4\right)-\left(x^2-16\right)\)

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

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

\(=\left(x+4\right)\left(6-x\right)\)

\(\text{d) }x^2+4x+3\)

\(=x^2+3x+x+3\)

\(=\left(x^2+3x\right)+\left(x+3\right)\)

\(=x\left(x+3\right)+\left(x+3\right)\)

\(=\left(x+3\right)\left(x+1\right)\)

\(\text{e) }x^2+5x-6\)

\(=x^2+6x-x-6\)

\(=\left(x^2+6x\right)-\left(x+6\right)\)

\(=x\left(x+6\right)-\left(x+6\right)\)

\(=\left(x+6\right)\left(x-1\right)\)