ĐKXĐ: a ≠0; b≠0; a và b cùng dấu
a/ \(A=\left[\left(\sqrt{\dfrac{a}{b}}-1\right)\left(\sqrt{\dfrac{b}{a}}+1\right)\right]:\left(\dfrac{a}{b}-\dfrac{b}{a}\right)\)
\(=\left[\left(\dfrac{\sqrt{a}}{\sqrt{b}}-1\right)\left(\dfrac{\sqrt{b}}{\sqrt{a}}+1\right)\right]:\left(\dfrac{a^2-b^2}{ab}\right)\)
\(=\left(\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{b}}\cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}}\right)\cdot\dfrac{ab}{a^2-b^2}\)
\(=\dfrac{a-b}{\sqrt{ab}}\cdot\dfrac{ab}{\left(a-b\right)\left(a+b\right)}=\dfrac{\sqrt{ab}}{a+b}\)
b/ A = 2 <=> \(\dfrac{\sqrt{ab}}{a+b}=2\)
\(\Leftrightarrow\dfrac{\sqrt{a}}{a+1}=2\Leftrightarrow\sqrt{a}=2\left(a+1\right)\)
\(\Leftrightarrow-2a+\sqrt{a}-2=0\)
Đặt \(\sqrt{a}=t\left(t>0\right)\)
pt <=> \(-2t^2+t-2=0\)
\(\Leftrightarrow-\left(2t^2-t+2\right)=0\)
\(\Leftrightarrow-\left[\left(2t^2-2\cdot\sqrt{2}\cdot t\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{15}{8}\right]=0\)
\(\Leftrightarrow-\left(\sqrt{2}\cdot t+\dfrac{\sqrt{2}}{4}\right)-\dfrac{15}{8}=0\)
Có: VT = \(-\left(\sqrt{2}t+\dfrac{\sqrt{2}}{4}\right)-\dfrac{15}{8}\le-\dfrac{15}{8}< 0\)
mà VP = 0
=> Pt vô nghiệm
=> Không có a nào thỏa mãn đề
vậy=-.
