\(a,\Leftrightarrow\left(x+3\right)^2-4\left(x-3\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-4x+12\right)=0\\ \Leftrightarrow\left(x+3\right)\left(15-3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

\(b,=x^2\left(y-1\right)-\left(y-1\right)^2=\left(y-1\right)\left(x^2-y+1\right)\)

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

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

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

b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)

d) x^6 + y^6 = (x^2)^3 + (y^2)^3 

= (x^2 + y^2)(X^2 - x^2.y^2 + y^2)

c) = (x+y)^3 + 3(x+y)^2z + 3((x+y)z^2 + z^3 - X^3 - Y^3 - z^3

= (x+y)^3 + 3(x+y)^2z + 3((x+y)z^2  - (x+y)(x^2 - xy + y^2)

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

= (X+y)(x^2 + 2xy + y^2 + 3xz + 3yz + 3z^2 - x^2 + xy - y^2)

= (x+y)(3xy + 3xz + 3z^2 + 3yz)

= (x+y)[3x(y+z) + 3z(y+z)]

=3(x+y)(y+z)(x+z)

Đúng thì  

= 

tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-2-y-2-3-z-2-x-2-3-y-2-z-2-3-thanh-nhan-tu-faq358831.html

Đây bạn nhé 

Bạn làm theo người này nhé, mik cũng khum bt. mik chỉ nhắc để bn có đáp án nhanh nhất thui 

1D  2C

Câu 1: D

Câu 2: C