Rút gọn biểu thức \(B=\left(\dfrac{4a-9a^{-1}}{2a^{\dfrac{1}{2}}-3a^{-\dfrac{1}{2}}}+\dfrac{a-4+3a^{-1}}{a^{\dfrac{1}{2}}-a^{-\dfrac{1}{2}}}\right)^2\) ta được
B=9a.B=3a.B=6a.B=12a.Hướng dẫn giải:\(B=\left(\dfrac{4a-9a^{-1}}{2a^{\dfrac{1}{2}}-3a^{-\dfrac{1}{2}}}+\dfrac{a-4+3a^{-1}}{a^{\dfrac{1}{2}}-a^{-\dfrac{1}{2}}}\right)^2\)
\(=\left(\dfrac{4a-9\dfrac{1}{a}}{2\sqrt{a}-3\dfrac{1}{\sqrt{a}}}+\dfrac{a-4+3\dfrac{1}{a}}{\sqrt{a}-\dfrac{1}{\sqrt{a}}}\right)^2\)
\(=\left(\dfrac{\dfrac{4a^2-9}{a}}{\dfrac{2a-3}{\sqrt{a}}}+\dfrac{\dfrac{a^2-4a+3}{a}}{\dfrac{a-1}{\sqrt{a}}}\right)^2\)
\(=\left(\dfrac{\dfrac{\left(2a-3\right)\left(2a+3\right)}{a}}{\dfrac{2a-3}{\sqrt{a}}}+\dfrac{\dfrac{\left(a-1\right)\left(a-3\right)}{a}}{\dfrac{a-1}{\sqrt{a}}}\right)^2\)
\(=\left(\dfrac{2a+3}{\sqrt{a}}+\dfrac{a-3}{\sqrt{a}}\right)^2\)
\(=\left(\dfrac{3a}{\sqrt{a}}\right)^2\)
\(=9a\)
