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1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
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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
b: \(=\left(x-y\right)^2-4y^2\)
\(=\left(x-y-2y\right)\left(x-y+2y\right)\)
\(=\left(x-3y\right)\left(x+y\right)\)
c: \(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
a) \(x^4+8x+63\)
\(=x^4+4x^3+9x^2-4x^3-16x^2-36x+7x^2+28x+63\)
\(=x^2\left(x^2+4x+9\right)-4x\left(x^2+4x+9\right)+7\left(x^2+4x+9\right)\)
\(=\left(x^2+4x+9\right)\left(x^2-4x+7\right)\)
c) \(\left(x^2+2x+7\right)+\left(x^2-2x+4\right)\left(x^2+2x+3\right)\left(1\right)\)
Ta có : \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
\(\Rightarrow x^2+2x+4=\dfrac{x^3-8}{x-2}\)
\(\left(1\right)\Rightarrow\left[\left(\dfrac{x^3-8}{x-2}+3\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-8}{x-2}-1\right)\right]\)
\(=\left[\left(\dfrac{x^3-3x-14}{x-2}\right)\right]+\left(x^2-2x+4\right)\left[\left(\dfrac{x^3-2x-5}{x-2}\right)\right]\)
\(=\dfrac{1}{x-2}\left[x^3-3x-14+\left(x^2-2x+4\right)\left(x^3-2x-5\right)\right]\)
a.
$64x^3-16x^2+x=x(64x^2-16x+1)$
$=x(8x-1)^2$
b.
$36-4xy+24y-x^2=(4y^2+24y+36)-(x^2+4xy+4y^2)$
$=(2y+6)^2-(x+2y)^2=(2y+6-x-2y)(2y+6+x+2y)$
$=(6-x)(x+4y+6)$
c.
$x^2+10x-2010.2020$
$=x^2+10x-(2015-5)(2015+5)
$=x^2+10x-(2015^2-5^2)$
$=(x^2+10x+5^2)-2015^2=(x+5)^2-2015^2$
$=(x+5-2015)(x+5+2015)=(x-2010)(x+2020)$
d.
$25x^2-121+22y-y^2$
$=(5x)^2-(y^2-22y+11^2)$
$=(5x)^2-(y-11)^2=(5x-y+11)(5x+y-11)$
e.
$(x^2+2x)(x^2+2x-2)-3$
$=(x^2+2x)^2-2(x^2+2x)-3$
$=(x^2+2x)^2+(x^2+2x)-3(x^2+2x)-3$
$=(x^2+2x)(x^2+2x+1)-3(x^2+2x+1)$
$=(x^2+2x+1)(x^2+2x-3)$
$=(x+1)^2[x(x-1)+3(x-1)]$
$=(x+1)(x-1)(x+3)$
\(=\left(x^2-2x+1\right)+2=\left(x-1\right)^2+2\)