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\(a,x^2+6x=x\left(x+6\right)\\ b,9x^2-1=\left(3x\right)^2-1^2=\left(3x-1\right)\left(3x+1\right)\\ c,x^2+2xy-9+y^2=\left(x^2+2xy+y^2\right)-9=\left(x+y\right)^2-3^2=\left(x+y-3\right)\left(x+y+3\right)\\ c,x^2-y^2-x+y=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
\(a,=5x\left(4x-1\right)\\ b,=y^2-\left(x-1\right)^2=\left(y-x+1\right)\left(y+x-1\right)\\ c,=6x^2+3x-4x-2=3x\left(x+2\right)-2\left(x+2\right)=\left(3x-2\right)\left(x+2\right)\)
a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
\(a,=x\left(2x-y\right)+\left(2x-y\right)=\left(x+1\right)\left(2x-y\right)\\ b,=\left(a+b\right)\left(c-2\right)\\ c,=x\left(x+4y\right)+2\left(x+4y\right)=\left(x+2\right)\left(x+4y\right)\\ d,=x\left(x+2y\right)+3\left(x+2y\right)=\left(x+3\right)\left(x+2y\right)\)
\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)
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Câu 2:
a: =x(x+6)
b: =(3x-1)*(3x+1)
c: \(=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
d: \(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
Câu a) dễ, ko làm
b) \(x^2y^2+1-x^2-y^2\)
\(=x^2\left(y^2-1\right)-\left(y^2-1\right)\)
\(=\left(x^2-1\right)\left(y^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
Câu c) đề sai
Câu c) ,đề đúng nek
\(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left[\left(b+c\right)-\left(a+b\right)\right]-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left(b+c\right)-ac\left(a+b\right)-ab\left(a+b\right)\)
\(=\left(b+c\right)\left(bc+ac\right)-\left(a+b\right)\left(ac+ab\right)\)
\(=\left(b+c\right)c\left(a+b\right)-\left(a+b\right)a\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\left(c-a\right)\)