1)

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)

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

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

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

c) \(\frac{x-6}{x+1}=\frac{x^2}{x-1}\)

nhân chéo lên, ngại chết đc

a)(x+1)^2

b)3x+y)^2

c)(5a-2b)^2

d)(x-1/2)^2

e)(3x-1)^2

\(9x^2-6x+1=\left(3x\right)^2-2.3x.1+1^2=\left(3x-1\right)^2\) (bình phương  1 hiệu)

\(\left(2x+3y\right)^2+2.\left(2x+3y\right)+1=\left(2x+3y\right)^2+2.\left(2x+3y\right)+1^2=\left(2x+3y+1\right)^2\) (bình phương 1 tổng)

a/ 9x²-12xy+4y² = (3x - 2y)²

b/ 25x²-10x+1 = (5x - 1)²

c/ 9x²-12x+4 = (3x - 2)²

d/ 4x²+20x+25 = (2x + 5)²

e/ x⁴-4x²+4 = (x² - 2)²

a/\(\left(3x-2y\right)^2\)

b/\(\left(5x-1\right)^2\)

c/\(\left(3x-2\right)^2\)

d/\(\left(2x+5\right)^2\)

e/\(\left(x-2\right)^2\)

để thằng này làm cho :

a/\(\left(3x\right)^2-12xy+\left(2y\right)^2\)

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

B/làm cx tương tự 

mấy câu kia cx z

a) \(2x^2+2b^2=x^2+b^2+x^2+b^2=x^2+2xb+b^2+x^2-2xb+b^2=\left(x+b\right)^2+\left(x-b\right)^2\)

                                                     Bài giải

Ta có :

\(1=1^2\) ; \(4=2^2\) ; \(9=3^2\)

\(1=1^3\) ; \(8=2^3\)

Đây nhoa : https://olm.vn/hoi-dap/question/101769.html