Đáp án:
Giải thích các bước giải:
`A=(a\sqrt{a}+b\sqrt{b})/[(\sqrt{a}+\sqrt{b})(a-b)]`
`B=\sqrt{ab}/(a-b) - (2\sqrt{b})/(a\sqrt{a}+b\sqrt{b})`
`B=[\sqrt{ab}(a\sqrt{a}+b\sqrt{b})-2\sqrt{b}(a-b)]/(a\sqrt{a}+b\sqrt{b})`
`B=[-a\sqrt{b} + b\sqrt{a}+2b\sqrt{b}]/[(a\sqrt{a}+b\sqrt{b})(a-b)]`
`=>A-B=(a\sqrt{a}+b\sqrt{b})/[(\sqrt{a}+\sqrt{b})(a-b)]-[-a\sqrt{b} + b\sqrt{a}+2b\sqrt{b}]/[(a\sqrt{a}+b\sqrt{b})(a-b)]`
`=>A-B=(a\sqrt{a}+b\sqrt{b}-(-a\sqrt{b} + b\sqrt{a}+2b\sqrt{b}))/[(\sqrt{a}+\sqrt{b})(a-b)]`
`=>A-B=(a\sqrt{a}+b\sqrt{b}+a\sqrt{b} - b\sqrt{a}-2b\sqrt{b})/[(\sqrt{a}+\sqrt{b})(a-b)]`
`=>A-B=[(\sqrt{a}+\sqrt{b})(a-b)]/[(\sqrt{a}+\sqrt{b})(a-b)]`
`=>A-B=1`
`=>đ.p.c.m`
