Bài 1:
\(a^2\left(b-2c\right)+b^2\left(c-a\right)+2c^2\left(a-b\right)+abc\)
\(=2c^2\left(a-b\right)+a^2b-ab^2+b^2c-a^2c+abc-a^2c\)
\(=2c^2\left(a-b\right)+ab\left(a-b\right)-c\left(a+b\right)\left(a-b\right)-ac\left(a-b\right)\)
\(=\left(a-b\right)\left(2c^2+ab-ac-cb-ac\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-2c\right)\)
Bài 2:
\(x^2+3x+1=0\Leftrightarrow x+\frac{1}{x}=-3\)(vì \(x=0\)không là nghiệm)
Ta có:
\(x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right).x.\frac{1}{x}=-3^3-3.\left(-3\right)=-18\)
\(x^4+\frac{1}{x^4}=\left(x^2+\frac{1}{x^2}\right)^2-2=\left[\left(x+\frac{1}{x}\right)^2-2\right]^2-2=47\)
\(\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)=x^7+\frac{1}{x^7}+x+\frac{1}{x}\)
\(\Leftrightarrow x^7+\frac{1}{x^7}=\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac{1}{x}\right)=-18.47-\left(-3\right)=-843\)
