e) Ta có: \(a^3-a^2-a+1\)
\(=a^2\left(a-1\right)-\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2-1\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)\)
f) Ta có: \(x^3-2xy-x^2y+2y^2\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)
b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
d) Đề sai ko ???
e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)
f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
a, \(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(\left(a-b\right)\left(a+b\right)\right)^2=\left(a^2-b^2\right)^2\)
\(b,=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
\(c,=-\left(x^2-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(d,=2\left(x^2+2xy+y^2-4z^2\right)=2\left(\left(x+y\right)^2-4z^2\right)=2\left(x+y-2z\right)\left(x+y+2z\right)\)
\(e,=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)\)
\(f,=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x^2-2y\right)\left(x-y\right)\)
a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)\)
\(=\left(a-b\right)^2\cdot\left(a+b\right)^2\)
b) Ta có: \(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
c) Ta có: \(-x^3+3x^2-3x+1\)
\(=1-3x+3x^2-x^3\)
\(=\left(1-x\right)^3\)
d) Ta có: \(2x^2+4xy+2y^2-8z^2\)
\(=2\left(x^2+2xy+y^2-4z^2\right)\)
\(=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)
\(=2\left(x+y-2z\right)\left(x+y+2z\right)\)
