Bài 2:
a) \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)\)
\(=a^3+b^3=VT\) (đpcm)
b) \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc\)\(+a^2c+b^2c+c^3-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc\)
Bài 1:
\(N=\frac{x\left|x-2\right|}{x^2+8x-20}+12x-3\)
\(=\frac{x\left|x-2\right|}{\left(x-2\right)\left(x+10\right)}+12x-3\)
Nếu \(x\ge2\)thì: \(N=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)
\(=\frac{x}{x+10}+12x+3\) (lm tiếp nhé)
Nếu \(x< 2\) thì: \(N=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)
\(=\frac{-x}{x+10}+12x-3\) (lm tiếp nhé)