b)Ta có: \(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow\frac{a^2}{ab}+\frac{b^2}{ab}\ge2\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}\ge2\left(đpcm\right)\)
\(a^5-a=a\left(a^4-1\right)\)
\(=a\left(a^2+1\right)\left(a^2-1\right)\)
\(=a\left(a^2+1\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a^2-4+5\right)\left(a-1\right)\left(a+1\right)\)
\(=a\left(a^2-4\right)\left(a-1\right)\left(a+1\right)+5a\left(a+1\right)\left(a-1\right)\)
\(=\left(a-2\right)\left(a-1\right)a\left(a+1\right)\left(a+2\right)+5a\left(a+1\right)\left(a-1\right)\)
Tích 5 số nguyên liên tiếp chia hết cho 5 nên \(a^5-a⋮5\)
Ta có:
\(\left(a+b\right)\left(ab+ac+bc+c^2\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[\left(ab+bc\right)+\left(ac+c^2\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
Không mất tính tổng quát giả sử \(a+b=0\Rightarrow a=-b\)
\(a+b+c=1\Rightarrow-b+b+c=1\Rightarrow c=1\)
Ta có:\(M=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{-1}{b^3}+\frac{1}{b^3}+\frac{1}{1^3}=1\)
Is that true ??
