\(8x^2-2\)

\(=2.\left(1^2-4x^2\right)\)

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

\(x^2-6x-y^2+9\)

\(=\left(x^2-6x+9\right)-y^2\)

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

\(=\left(x-3-y\right).\left(x-3+y\right)\)

Mình sửa lại câu đầu ạ.

\(8x^2-2\)

\(=2.\left(4x^2-1\right)\)

\(=2.[\left(2x\right)^2-1^2]\)

\(=2.\left(2x-1\right).\left(2x+1\right)\)

a) \(x^2-8y^2+6x+9\)

\(=\left(x^2+6x+9\right)-8y^2\)

\(=\left(x+3\right)^2-\left(\sqrt{8}\cdot y\right)^2\)

\(=\left(x+3+\sqrt{8}y\right)\left(x+3-\sqrt{8}y\right)\)

Câu 3: 

a: 2x-8=4

nên 2x=12

hay x=6

b: 7x-3x=2x+7

\(\Leftrightarrow4x-2x=7\)

hay \(x=\dfrac{7}{2}\)

Câu 1:

a: \(5x\left(3x-4\right)=15x^2-20x\)

b: \(\left(x+5\right)\left(x-5\right)=x^2-25\)

c) \(5x^2.\left(3x^2-7x+2\right)\)

\(=5x^2.3x^2-5x^2.7x+5x^2.2\)

\(=15x^4-35x^3+10x^2\)

d) \(\dfrac{3x^2y^2+6x^2y^3-12xy}{3xy}\)

\(=\dfrac{3xy\left(xy+2y-4\right)}{3xy}\)

\(=xy+2y-4\)

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(x^3+y^3-3x^2+3x-1\\=(x^3-3x^2+3x-1)+y^3\\=(x-1)^3+y^3\\=(x-1+y)[(x-1)^2-(x-1)y+y^2]\\=(x+y-1)(x^2-2x+1-xy+y+y^2)\)

\(x^3-3x^2y+x+3xy^2-y-y^3\\=(x^3-3x^2y+3xy^2-y^3)+(x-y)\\=(x-y)^3+(x-y)\\=(x-y)[(x-y)^2+1]\\=(x-y)(x^2-2xy+y^2+1)\)

 \(\left(x-y\right)^5+\left(y-z\right)^5+\left(z-x\right)^5\)

\(=x^5-5x^4y+10x^3y^2-10x^2y^3+5xy^4-y^5+y^5-5y^4z+10y^3z^2-10y^2z^3+5yz^4-z^5\)\(+z^5-5z^4x+10z^3x^2-10z^2x^3+5zx^4-x^5\)

\(=5\left(-x^4y+2x^3y^2-2x^2y^3+xy^4-y^4z+2y^3z^2-2y^2z^3+yz^4-z^4x+2z^3x^2-2z^2x^3+zx^4\right)\)