\(4x^2-y^2+10y-25\)
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\(=x^2-2xy-5xy+10y^2=x\left(x-2y\right)-5y\left(x-2y\right)=\left(x-2y\right)\left(x-5y\right)\)
\(4x^2-9y^2+4x-6y=\left(4x^2-9y^2\right)+\left(4x-6y\right)=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12=\left[\left(x^2+x\right)^2+4\left(x^2+x\right)+4\right]-16=\left(x^2+x+2\right)-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
\(\left(x^2+x\right)^2+4x^2+4x-12\\ =\left(x^2+x+2\right)-4\\ =\left(x^2+x-2\right)\left(x^2+x+6\right)\)
(1 + x2)2 - 4x(1 - x2)
= (1 + x2)(1 + x2) - 4x(1 - x2)
= (1 + x2 - 4x)(1 + x2 - 1 + x2)
= 2x2(x2 - 4x + 1)
Ta có: \(\left(x^2+1\right)^2+4x\left(x^2-1\right)\)
\(=x^4+2x^2+1+4x^3-4x\)
\(=x^4+2x^3+2x^3+4x^2-2x^2-4x+1\)
\(=\left(x+2\right)\left(x^3+2x^2-2x\right)+1\)
\(=\left(x^2+5x+8\right)\left(x^2+4x+2x+8\right)=\left(x^2+5x+8\right)\left[x\left(x+4\right)+2\left(x+4\right)\right]\)
\(=\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)\)
\(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=\left(x^2+4x+8\right)^2+2x\left(x^2+4x+8\right)+x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+2x\right)+x\left(x^2+4x+8+2x\right)\)
\(=\left(x^2+4x+8\right)\left(x^2+6x+8\right)+x\left(x^2+6x+8\right)\)
\(=\left(x^2+4x+8+x\right)\left(x^2+6x+8\right)=\left(x^2+5x+8\right)\left(x^2+6x+8\right)\)
4x^2 − y^2 + 10y − 25
= (2x)^2 − ( y^2 − 2.y.5 + 5^2)
= (2x)^2− (y−5)^2
= ( 2x−y+5 ) ( 2x+y−5 )