Question

Hộ mik vớiiii

Answer 1

a)ĐKXĐ:\(x\ge0,x\ne4\)

b)M=\(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\dfrac{x-4}{\sqrt{4x}}\)

M=\(\dfrac{2x}{x-4}.\dfrac{x-4}{\sqrt{4x}}\)=\(\sqrt{x}\)

c) để M>3\(\Rightarrow\)\(\sqrt{x}>3\)\(\Rightarrow x>9\)

Answer 2

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\cdot\dfrac{x-4}{\sqrt{4x}}\)

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

\(=\sqrt{x}\)

c: Để M>3 thì x>9