bạn có thể ghi rõ lại được ko ạ, bạn có thể thay dấu nhân bằng dấu chấm

Ta có: \(3x\left(x-1\right)^2-5x^2\left(x-1\right)-7\left(x-1\right)\)

\(=\left(x-1\right)\left[3x\left(x-1\right)-5x^2-7\right]\)

\(=\left(x-1\right)\left(3x^2-3x-5x^2-7\right)\)

\(=\left(x-1\right)\left(-2x^2-3x-7\right)\)

\(=3x\cdot\left(x-5y\right)+2y\cdot\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2y\right)\)

\(3x\left(x-5y\right)-2y\left(5y-x\right)\)

\(=3x\left(x-5y\right)+2y\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2y\right)\)

`3x-3y+a(x-y)`

`=3(x-y)+a(x-y)`

`=(x-y)(a+3)`

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

\(3x\left(x-7\right)+2xy-14y\)

\(=3x\left(x-7\right)+2y\left(x-7\right)\)

\(=\left(x-7\right)\left(3x+2y\right)\)

#\(Toru\)

3x² - 3xy

= 3x ( x - y )

`3x^2 - 6x`

`= 3x.x-3x.2`

`=3x(x-2)`

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

Cưu là mình vs (x^2+x)^2-2(x^2+x)-15

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)