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https://olm.vn/bg/luyenthichuyen/thao-luan

Ta có: \(\left(\sqrt{a}+\sqrt{c}\right)^2=a+2\sqrt{ac}+c=2b+2\sqrt{ac}\)(1)

Lại có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2\sqrt{b}+\sqrt{a}+\sqrt{c}}{b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}\)

\(=\frac{\left(2\sqrt{b}+\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(Nhân cả tử & mẫu với \(\sqrt{a}+\sqrt{c}\))

\(=\frac{2\sqrt{ab}+2\sqrt{bc}+\left(\sqrt{a}+\sqrt{c}\right)^2}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)(2)

Thế (1) và (2) => \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}\)\(=\frac{2\sqrt{ab}+2\sqrt{bc}+2b+\sqrt{ca}}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}=\frac{2\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}{\left(b+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\left(\sqrt{a}+\sqrt{c}\right)}\)

\(=\frac{2}{\sqrt{a}+\sqrt{c}}.\)

\(\Rightarrow\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\)(đpcm).

kurokawa neko sau khi thay 1 vào 2 là 2\(\sqrt{ac}\)nha

ta có \(\sqrt{\frac{a}{1-a}}=\frac{a}{\sqrt{a\left(1-a\right)}}\)

áp dụng cô si

\(\sqrt{a\left(1-a\right)}< =\frac{a+1-a}{2}=\frac{1}{2}\)

do do\(\sqrt{\frac{a}{1-a}}>=2a\)\(\sqrt{\frac{b}{1-b}}>=2b,\sqrt{\frac{c}{1-c}}>=2c\)

cmtt\(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}>=2\left(a+b+c\right)=2\left(doa+b+c=1\right)\)

dau = xay ra <=>\(\hept{\begin{cases}a=1-a\\b=1-b\\c=1-c\end{cases}=>a+b+c=3-\left(a+b+c\right)}\)

<=>2(a+b+c)=3

<=>a+b+c=3/2 

vay dau = khong xay ra ta co dpcm

bn j ui mk tưởng cs TH = 2 nx mà...