Question

A=\(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\) biết a=\(\sqrt{4+2\sqrt{3}}\) , b=\(\sqrt{4-2\sqrt{3}}\)

Answer 1

Ta có : \(a=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Tương tự ta có \(b=\sqrt{3}-1\)

Thiết lập được : \(\sqrt{ab}=\sqrt{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}=\sqrt{3-1}=\sqrt{2}\)

\(a+b=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

Khi đó : \(A=\frac{\sqrt{3}+1}{\sqrt{2}+\sqrt{3}-1}+\frac{\sqrt{3}-1}{\sqrt{2}-\sqrt{3}-1}-\frac{2\sqrt{3}}{\sqrt{2}}\)

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