\(=\dfrac{2\left(x-2y+z\right)^3+4\left(x-2y+z\right)^2}{2\left(x-2y+z\right)}=\left(x-2y+z\right)^2+2\left(x-2y+z\right)\)

Anh chỉ em pp học Toán giỏi như anh được ko ạ  rất thích môn Toán nên cũng rất hâm mộ anh

a) 2x(y-z)-6y(z-y)

=2x(y-z)+6y(y-z)

=2(y-z)(x+3y)
b)x^2+4x-4y-y^2

=x^2-y^2+4x-4y

=(x-y)(x+y)+4(x-y)

=(x-y)(x+y+4)

P/s tham khảo nha

a)  = 2x(y-z)-6y(y-z)

= 2(y-z)(x-3y)

b)  = 

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

a) Ta có: \(4x\left(2y-z\right)+7y\left(z-2y\right)\)

        \(=4x\left(2y-z\right)-7y\left(2y-z\right)\)

        \(=\left(4x-7y\right)\left(2y-z\right)\)

b) Ta có: \(2x\left(x+3\right)+\left(3+x\right)\)

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

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

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

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

^^

Gia Đình Là Vô Giá

dòng 2 bước chuyển đổi chưa hiểu lắm!

sao đùng 1 cái lại mất x của em vậy ?

1.A

2.C

3.B

4.C

a

c

b

c

\(8x^3\left(y+z\right)-y^3\left(z+2x\right)-z^3\left(2x-y\right)\)

\(=8x^3\left(y+z\right)-y^3\left[\left(y+z\right)+\left(2x-y\right)\right]-z^3\left(2x-y\right)\)

\(=8x^3\left(y+z\right)-y^3\left(y+z\right)-y^3\left(2x-y\right)-z^3\left(2x-y\right)\)

\(=\left(y+z\right)\left(8x^3-y^3\right)-\left(2x-y\right)\left(y^3+z^3\right)\)

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

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

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