Question

giai chi tiet cho minh nha mn ^^:

^{cho a,b,c>0 thõa mãn 2b=a+c. Cmr: \(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{2}{\sqrt{c}+\sqrt{a}}\)}

Answer 1

Qui đồng thôi :|

\(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{\sqrt{a}+\sqrt{c}+2\sqrt{b}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}+b}\)

Thay \(b=\dfrac{a+c}{2}\) vào cái mẫu:

\(M=\dfrac{1}{2}\left(2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}+a+c\right)\)

\(=\dfrac{1}{2}\left(2\sqrt{ab}+\sqrt{ac}+a\right)+\dfrac{1}{2}\left(c+\sqrt{ac}+2\sqrt{bc}\right)\)( nhóm tách sao cho xuất hiện tử)

\(=\dfrac{1}{2}\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}+2\sqrt{b}\right)\)

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