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ĐK: a>0,b>0,a\(\ne b\)
a) \(K=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}+\frac{\sqrt{b}}{\sqrt{ab}-a}\right).\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left(\frac{\sqrt{a}}{\sqrt{ab}-b}-\frac{\sqrt{b}}{a-\sqrt{ab}}\right).\frac{\sqrt{a}+\sqrt{b}}{a\sqrt{b}-b\sqrt{a}}=\left[\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right].\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}=\frac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{ab\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2\left(\sqrt{a}-\sqrt{b}\right)}{ab\left(\sqrt{a}-\sqrt{b}\right)^2}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{ab\left(\sqrt{a}-\sqrt{b}\right)}\)
b) Thay a=\(4+2\sqrt{3}\) và \(b=4-2\sqrt{3}\) vào K thì \(K=\frac{\left(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\right)^2}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)}=\frac{\left[\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right]^2}{\left(16-12\right)\left[\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\right]}=\frac{\left(\sqrt{3}+1+\sqrt{3}-1\right)^2}{4.\left(\sqrt{3}+1-\sqrt{3}+1\right)}=\frac{\left(2\sqrt{3}\right)^2}{8}=\frac{12}{8}=\frac{3}{2}\)
a) B= \(\frac{1}{\sqrt{a}}\)(ĐKXĐ: a,b>0) B) Khi a= \(6+2\sqrt{5}\)thì B=\(\frac{1}{\sqrt{\left(\sqrt{5}+1\right)^2}}\)=\(\frac{1}{\sqrt{5}+1}\) C) Do \(\sqrt{a}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>-1\)
