NQ

giúp mk vs

NT
17 tháng 8 2023 lúc 20:59

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

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\)

b: \(H=\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+4\sqrt{x}-4}{x-4}\right):\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}\)

\(=\dfrac{x+5\sqrt{x}+6+x-3\sqrt{x}+2+4\sqrt{x}-4}{x-4}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(=\dfrac{2x+6\sqrt{x}+4}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+3}\)

c: \(I=\dfrac{3x+3-2\sqrt{x}\left(\sqrt{x}-3\right)-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3\sqrt{x}+3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)

d: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}\right)\)

\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}{25-x+x-25-x+9}\)

\(=\dfrac{-5}{1}\cdot\dfrac{\sqrt{x}-3}{-x+9}\)

\(=\dfrac{5\left(\sqrt{x}-3\right)}{x-9}=\dfrac{5}{\sqrt{x}+3}\)

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