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manh

a : \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)với x ≥ 0 x ≠ 9

b : \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\)với x ≥ 0 x ≠ 1

c : \(\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)với x ≥ 0 x ≠ 0

d : \(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)với x ≥ 0 x ≠ 1

_gialinh.2901
7 tháng 10 2023 lúc 16:05

a) \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(x\ge0;x\ne0\right)\)

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

\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x-3}\right)}\)

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

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

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

_gialinh.2901
7 tháng 10 2023 lúc 16:26

b) \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\left(x\ge0;x\ne1\right)\)

\(=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

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

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

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

_gialinh.2901
7 tháng 10 2023 lúc 16:32

c) \(\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne1\right)\)

\(=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}+\dfrac{2.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

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

_gialinh.2901
7 tháng 10 2023 lúc 16:40

d) \(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\left(x\ge0;x\ne1\right)\)

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

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

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

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

\(=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+3\right)}\)

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


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