1) \(M=\dfrac{\dfrac{a^3}{b^3}-\dfrac{b^3}{a^3}+\dfrac{a^2-b^2}{ab}}{\left(\dfrac{a^2}{b^2}-\dfrac{b^2}{a^2}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\) (Đk: \(a,b>0;a\ne b\))
\(=\dfrac{\left(\dfrac{a}{b}-\dfrac{b}{a}\right)\left(\dfrac{a^2}{b^2}+1+\dfrac{b^2}{a^2}\right)+\dfrac{a}{b}-\dfrac{b}{a}}{\left(\dfrac{a}{b}-\dfrac{b}{a}\right)\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\)
\(=\dfrac{\left(\dfrac{a}{b}-\dfrac{b}{a}\right)\left(\dfrac{a^2}{b^2}+2+\dfrac{b^2}{a^2}\right)}{\left(\dfrac{a}{b}-\dfrac{b}{a}\right)\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\)
\(=\dfrac{\left(\dfrac{a}{b}+\dfrac{b}{a}\right)^2}{\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}\)
\(=\dfrac{\dfrac{a}{b}+\dfrac{b}{a}}{\dfrac{1}{a}+\dfrac{1}{b}}=\dfrac{a^2+b^2}{a+b}\)
2) Ta có: \(a+b=2\)
\(\Rightarrow a^2+b^2=a^2+\left(2-a\right)^2=2\left(a^2-2a+2\right)=2\left(a-1\right)^2+2\)
Vì a,b là các số dương phân biệt và a+b=2 nên a,b không thể đồng thời cùng bằng 1. Do đó \(\left(a-1\right)^2>0\)
\(\Rightarrow a^2+b^2=2\left(a-1\right)^2+2>2\)
Khi đó \(M=\dfrac{a^2+b^2}{a+b}=\dfrac{a^2+b^2}{2}>\dfrac{2}{2}=1\)
