Question

Cho biểu thức .

Q =\(\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)

a ) Tìm x để Q có nghĩa.

b) Rút gon Q .

Answer 1

a) điều kiện xác định \(x>0;x\ne9;x\ne4\)

b) ta có : \(Q=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)

\(\Leftrightarrow Q=\left(\dfrac{\sqrt{x}-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right):\left(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(\Leftrightarrow Q=\left(\dfrac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right):\left(\dfrac{x-9-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(\Leftrightarrow Q=\left(\dfrac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right):\left(\dfrac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\) \(\Leftrightarrow Q=\left(\dfrac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\right)\) \(\Leftrightarrow Q=\dfrac{3\left(\sqrt{x}-2\right)}{-5\sqrt{x}}\)

Answer 2

Câu a : ĐKXĐ : \(x>0\) ; \(x\ne4\) và \(x\ne9\)

\(Q=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\)

\(=\dfrac{\sqrt{x}-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}:\dfrac{-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-3\right)}\times\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{-5}\)

\(=\dfrac{3\left(\sqrt{x}-2\right)}{-5\sqrt{x}}\)

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