Question

\(A=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)

Với x > 0 ,x # 2

 \(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

Với a > 0 ,a # 4

Rút gọn A,M :

Answer 1

a: \(A=\dfrac{x\sqrt{2}}{x\sqrt{2}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{x-2}\)

\(=\dfrac{1}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+\sqrt{2}}\)

b: \(M=\left(\dfrac{\sqrt{a}+a}{\sqrt{a}-2}\right)\cdot\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\cdot\left(\sqrt{a}-2\right)=\sqrt{a}\left(\sqrt{a}-2\right)\)

Answer 2

\(A=\dfrac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\dfrac{\sqrt{2x}-2}{x-2}\)

\(=\dfrac{\sqrt{x}.\sqrt{2x}}{\sqrt{2x}\left(\sqrt{x}+\sqrt{2}\right)}+\dfrac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}+\dfrac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}=1\)

\(M=\left(\dfrac{a}{a-2\sqrt{a}}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{a-4\sqrt{a}+4}\)

\(=\left(\dfrac{a}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{a}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-2\right)^2}\)

\(=\dfrac{\sqrt{a}+a}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-2}.\dfrac{\left(\sqrt{a}-2\right)^2}{\sqrt{a}+1}\)

\(=\sqrt{a}\left(\sqrt{a}-2\right)\)