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\(=\left(3x+15\right)^2-\left(x-7\right)^2\\ =\left(3x+15-x+7\right)\left(3x+15+x-7\right)\\ =\left(2x+22\right)\left(4x+8\right)\\ =2\left(x+11\right)4\left(x+2\right)\\ =8\left(x+11\right)\left(x+2\right)\)
\(=\left(x^2+y^2-5\right)^2-\left[2\left(xy+2\right)\right]^2\\ =\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\\ =\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\\ =\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left[\left(x-y\right)^2-9\right]\left[\left(x-y\right)^2-1\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x-y-1\right)\left(x-y+1\right)\)
= \(-\left(x^2+4xy+4y^2\right)\)
= \(-\left(x+2y\right)^2\)
\(x^3-4x\)
\(=x\cdot\left(x^2-4\right)\)
\(=x\cdot\left(x^2-2^2\right)\)
\(=x\left(x+2\right)\left(x-2\right)\)
\(x^2-y^2-2x-2y=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
h: =3(x^2+3x-10)
=3(x^2+5x-2x-10)
=3[x(x+5)-2(x+5)]
=3(x+5)(x-2)
e: =x^2-2x-4x+8
=x(x-2)-4(x-2)
=(x-2)(x-4)
h: =x^2-3x-4x+12
=x(x-3)-4(x-3)
=(x-3)(x-4)
Lời giải:
b. $3x^2+9x-30=3(x^2+3x-10)$
$=3[(x^2-2x)+(5x-10)]=3[x(x-2)+5(x-2)]=3(x-2)(x+5)$
e. $x^2-6x+8=(x^2-2x)-(4x-8)=x(x-2)-4(x-2)=(x-4)(x-2)$
h. $x^2-7x+12=(x^2-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)$