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

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

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

\(3xy+3y+2x+2\)

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

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

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

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

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

a, \(x^3+y^3+3xy=\left(x+y\right)\left(x^2-xy+y^2\right)+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=\left(x+y\right)^2=1\)

b, tương tự a

c, Sửa đề Cho a+b=1. Tính giá trị của các biểu thứ :A= a³+b³+3ab(a²+b²)+ 6a²b²(a+b)

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2\left(a+b\right)\)

Thay a+b=1 vào A ta có:

\(A=1-3ab+3ab\left(1-2ab\right)+6a^2b^2\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2=1\)

d. \(B=x^2+2xy+y^2-4x-4y+1=\left(x+y\right)^2-4\left(x+y\right)+1=\left(x+y\right)\left(x+y-4\right)+1\)

Thay x+y=3 vào B ta có:

\(B=3\left(3-4\right)+1=3.\left(-1\right)+1=-3+1=-2\)

a) y² + 2y = y(y + 2)

b) y³ - 2y² + y = y(y² - 2y + 1) = y(y - 1)²

c) y² - x² - 6y - 6x 

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

<=> (x + y)( y - x - 6)

d) x³ - 3x = x(x² - 3)

e) 2x - xy + 2z - yz 

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

= (2 - y)(x + z)

\(\hept{\begin{cases}x^3+16x=6x^2+9\\9y^2+32=y^2+31y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^3-6x^2+16x-9=0\\9y^2-y^2-31y+32=0\end{cases}}\)

Đề sai sao ý 

đề bài đúng nhé, mak mk cũng lm đc rồi

a) \(x^3-4x^2-8x+8\)

\(=x^3+8-4x^2-8x\)

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

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

\(=\left(x+2\right)\left(x^2-6x+4\right)\)

\(=\left(x+2\right)\left(x^2-6x+9-5\right)\)

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

\(=\left(x+2\right)\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)

b) \(1+6x-6x^2-x\)

\(=1-x+6x\left(1-x\right)\)

\(=\left(1-x\right)\left(6x+1\right)\)

Ta có: a+b=-5

<=>(a+b)²=25

<=>a²+2ab+b²=25

<=>a²+b²+12=25

<=>a²+b²=17

Ta có: \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Thay a+b=-5,ab=6,a²+b²=17 vào biểu thức trên ta được:

\(-5\left(17-6\right)=-5.11=-55\)

ta có

\(a+b=5\)

=>\(\left(a+b\right)^2=25\)

=>\(a^2+2ab+b^2=25\)

=>\(a^2+2.6+b^2=25\)

=>\(a^2+12+b^2=25\)

=>\(a^2+b^2=17\)

ta có

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

thay \(a+b=-5;ab=6;a^2+b^2=17\) vào bt trên ,ta có

\(-5\left(17-6\right)=-5.11=-55\)

vậy \(a^3+b^3=-55\)