ta có :
\(A=\frac{a}{\sqrt{a}-1}-\frac{2a-\sqrt{a}}{a-\sqrt{a}}=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}=\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\sqrt{a}-1\)
mà \(a=3-\sqrt{8}=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\)
\(\Rightarrow\sqrt{a}=\sqrt{2}-1\Rightarrow A=\sqrt{2}-1-1=\sqrt{2}-2\)
ĐK : a > 0 , a khác 1
\(A=\frac{a}{\sqrt{a}-1}-\frac{\sqrt{a}\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{a}{\sqrt{a}-1}-\frac{2\sqrt{a}-1}{\sqrt{a}-1}\)
\(=\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\)
Với \(a=3-\sqrt{8}\left(tmđk\right)\)thay vào A ta được :
\(A=\sqrt{3-\sqrt{8}}-1=\sqrt{\left(\sqrt{2}-1\right)^2}-1=\left|\sqrt{2}-1\right|-1=\sqrt{2}-1-1=\sqrt{2}-2\)
