a) 5x(x-1)+10x-10=0

5x(x-1)+10(x-1)=0

(x-1)(5x+10)=0

\(\orbr{\begin{cases}x-1=0\\5x+10=0\end{cases}}\)

\(\orbr{\begin{cases}x=1\\5x=-10\end{cases}}\)

\(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

bài 2:

a, 5x( x-1 ) + 10x -10 =0

5x( x-1 ) + 10.( x-1 ) =0

( x-1 ) ( 5x + 10 ) =0

\(\Rightarrow\orbr{\begin{cases}x-11=0\\5x+10=0\end{cases}}\)=> \(\orbr{\begin{cases}x=11\\5x=-10\end{cases}}\) => \(\orbr{\begin{cases}x=11\\x=-2\end{cases}}\)

vậy x= 11 ; x=-2

b, ( x +2 ) ( x+3 ) -2x =6

(x+2) ( x+3) -2x -6 =0

(x +2)(x +3)-2(x+3)=0

(x+3) ( x+2-2)=0 => x(x+3)=0

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

vậy x=0; x=-3

c, ( x-1 ) ( x-2 ) -2 =0

x² - 2x -x +2 -2 =0

x² - x =0

x(x -1)=0

=> \(\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\) => \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

vậy x=0; x=1

Bài 2.

\(a.-x^3+3x^2-3x+1=\left(-x+1\right)^3\)    \(b.x^3+x^2+\frac{x}{3}+\frac{1}{27}=\left(x+\frac{1}{3}\right)^3\)

\(c.8-12x+6x^2-x^3=\left(2-x\right)^3\)     \(d.x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

bài 3.

\(A=x^3+3x^2y+3xy^3+y^3-\left(x^3-3x^2y+3xy^3-y^3\right)-2y^3=6x^2y\)

\(B=\left(x^3-6x^2+12x-8\right)-\left(x^3-x\right)+6x^2-18x=-5x-8\)

Bài 4.

\(a.\left(x+1\right)^3-x^2\left(x+3\right)=x^3+3x^2+3x+1-x^3-3x^2=3x+1=2\Leftrightarrow x=\frac{1}{3}\)

\(b.\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)\)\(=12x-4=-10\Leftrightarrow x=-\frac{1}{2}\)