Đáp án:
\(P=\dfrac{1}{ab}\)
Giải thích các bước giải:
Ta có:
`a^2/b^2+b^2/a^2-(a/b+b/a)`
`=a^2/b+2+b^2/a^2-(a/b+b/a)-2`
`=(a/b+b/a)^2-(a/b+b/a)-2`
`=(a/b+b/a)^2-2(a/b+b/a)+a/b+b/a-2`
`=(a/b+b/a)(a/b+b/a-2)+a/b+b/a-2`
`=(a/b+b/a-2)(a/b+b/a+1)`
\(\Rightarrow P=\dfrac{\left(\dfrac{a}{b}+\dfrac{b}{a}+1\right)\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2}{\left(\dfrac{a}{b}+\dfrac{b}{a}+1\right)\left(\dfrac{a}{b}+\dfrac{b}{a}-2\right)}\)
\(P=\dfrac{\left(\dfrac{1}{a}-\dfrac{1}{b}\right)^2}{\dfrac{a}{b}+\dfrac{b}{a}-2}\)
\(P=\dfrac{\left(\dfrac{b-a}{ab}\right)^2}{\dfrac{a^2-2ab+b^2}{ab}}\)
\(P=\dfrac{\dfrac{(a-b)^2}{(ab)^2}}{\dfrac{(a-b)^2}{ab}}\)
\(P=\dfrac{1}{ab}\)
