a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

x⁸ + 2500

= (x⁴)² + 2.x⁴.50 + (50)² - 2.x⁴.50

= (x⁴ + 50)² - (10x²)²

= (x⁴ + 50 - 10x²)(x⁴ + 50 + 10x²)

      \(X^2-2XY+Y^2+2X-2Y\)

\(\Leftrightarrow\left(X^2-2XY+Y^2\right)+\left(2X-2Y\right)\)

\(\Leftrightarrow\left(X-Y\right)^2+2\left(X-Y\right)\)

\(\Leftrightarrow\left(X-Y\right)\left(X-Y+2\right)\)

Tk mình nhé.

\(X^2\) là mũ 2 à bạn? 

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

2x²+x-3

=2x²+3x-2x-3

=(2x²-2x)+(3x-3)

=2x(x-1)+3(x-1)

=(x-1)(2x+3)

a) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)

\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)

\(\Leftrightarrow x-1.5=0\)

hay x=1,5

b) Ta có: \(2x^3+3x^2+2x+3=0\)

\(\Leftrightarrow\left(2x+3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

x(x +2y)^3-y(2x+y)^3 = [x(2x-y)-y(2x+y)].[x²(x +2y)² + x(x +2y).y(2x+y) + y²(2x+y)²

                                    = (2x-y)

Ta có: \(\left(x-2\right)^3+\left(5-2x\right)^3=0\)

\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)

\(\Leftrightarrow3-x=0\)

hay x=3

\(x^3+2x^2-2x-12=x^3-2x^2+4x^2-8x+6x-12\)

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

\(x^3+2x^2-2x-12\)

\(=x^3-2x^2+4x^2-8x+6x-12\)

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

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

hk tốt

^^