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tao biết làm bài này từ lớp 7 rồi, lớp 9 cũng hỏi mấy câu này
Đặt \(A=\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\left(\sqrt{b}+\sqrt{c}\right)}{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}\right)\left(\sqrt{a}+\sqrt{d}-\left(\sqrt{b}+\sqrt{c}\right)\right)}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{\left(\sqrt{a}+\sqrt{d}\right)^2-\left(\sqrt{b}+\sqrt{c}\right)^2}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+2\sqrt{ad}+d-\left(b+2\sqrt{bc}+c\right)}\)
Mà \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow ad=bc\)
\(\Rightarrow A=\frac{\sqrt{a}-\sqrt{b}-\sqrt{c}+\sqrt{d}}{a+2\sqrt{bc}+d-b-2\sqrt{bc}-c}\)
\(=\frac{\sqrt{a}-\sqrt{b}-\sqrt{c}+\sqrt{d}}{a-b-c+d}\)
\(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\dfrac{5-2\sqrt{5}+1}{5-1}=\dfrac{2\left(3-\sqrt{5}\right)}{4}=\dfrac{3-\sqrt{5}}{2}\)
b: \(\dfrac{37}{7+2\sqrt{3}}=7-2\sqrt{3}\)
c:\(=\dfrac{\sqrt{5}\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\left(2\sqrt{2}-\sqrt{5}\right)}=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\)
d: \(=\dfrac{\left(1+\sqrt{a}\right)\cdot\left(2+\sqrt{a}\right)}{4-a}\)
Ta có : \(ad=bc;a,b,c,d>0\)
\(\Rightarrow2\sqrt{ad}=2\sqrt{bc}\)
Khi đó : \(\frac{1}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}\) \(=\frac{1}{\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)}{\left[\left(\sqrt{a}+\sqrt{d}\right)+\left(\sqrt{b}+\sqrt{c}\right)\right].\left[\left(\sqrt{a}+\sqrt{d}\right)-\left(\sqrt{b}+\sqrt{c}\right)\right]}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{\left(\sqrt{a}+\sqrt{d}\right)^2-\left(\sqrt{b}+\sqrt{c}\right)^2}\) \(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d+2\sqrt{ad}-b-c-2\sqrt{bc}}\)
\(=\frac{\sqrt{a}+\sqrt{d}-\sqrt{b}-\sqrt{c}}{a+d-b-c}\) ( Do \(2\sqrt{ad}=2\sqrt{bc}\) )