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thu dinh
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Nguyễn Lê Phước Thịnh
3 tháng 3 2021 lúc 13:51

Ta có: \(P=\left(\dfrac{3x-6\sqrt{x}}{x\sqrt{x}-2x}-\dfrac{1}{2-\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}\right)\cdot\left(1-\dfrac{\sqrt{x}-3}{\sqrt{x}-2}\right)\)

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

\(=\left(\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-2}\)\(=\dfrac{3\sqrt{x}-6+\sqrt{x}+x-5\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}-2}\)

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

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

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

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thu dinh
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Nguyễn Lê Phước Thịnh
3 tháng 3 2021 lúc 13:54

Ta có: \(P=\left(1-\dfrac{x-3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)

\(=\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right)\)

\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{-\left(x-9\right)+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\dfrac{\sqrt{x}+3-\sqrt{x}}{\sqrt{x}+3}:\dfrac{-x+9+2x-4\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

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

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

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thu dinh
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NLT MInh
2 tháng 3 2021 lúc 11:22

ĐKXĐ x\(\ge0,x\ne1,x\ne4\)

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

 

P=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{x+2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)

P=\(\dfrac{x-1+\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{3\left(x-3\right)}\)

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

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thu dinh
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Đức Hiếu
3 tháng 3 2021 lúc 11:04

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

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

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

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

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Tuyết Linh Linh
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Đức Hiếu
2 tháng 3 2021 lúc 9:40

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

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

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

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

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NLT MInh
2 tháng 3 2021 lúc 10:57

ĐKXĐ x\(\ge0;x\ne1\)

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

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

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hello hello
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Thiên Thương Lãnh Chu
18 tháng 2 2021 lúc 16:20

P = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)DKXD: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

   = \(\sqrt{x}+\sqrt{x}\)

   = \(2\sqrt{x}\)

Vậy tại x ∈ ĐKXĐ thì P = \(2\sqrt{x}\)

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Nguyễn Lê Phước Thịnh
18 tháng 2 2021 lúc 22:54

Ta có: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

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

\(=2\sqrt{x}\)

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Trang Nguyễn
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Akai Haruma
26 tháng 8 2021 lúc 13:10

Lời giải:
ĐKXĐ: $x>0$

a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)

\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\frac{(\sqrt{x}+1)^2}{\sqrt{x}}\)

b.

\(x=\frac{4}{4+2\sqrt{3}}=(\frac{2}{\sqrt{3}+1})^2\Rightarrow \sqrt{x}=\frac{2}{\sqrt{3}+1}\)

\(P=\frac{(\frac{2}{\sqrt{3}+1}+1)^2}{\frac{2}{\sqrt{3}+1}}=\frac{3+3\sqrt{3}}{2}\)

 

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Nguyễn Lê Phước Thịnh
26 tháng 8 2021 lúc 13:16

a: Ta có: \(P=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

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

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

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

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thu dinh
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Nguyễn Lê Phước Thịnh
3 tháng 3 2021 lúc 13:45

Ta có: \(P=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)

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

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

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

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thu dinh
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Nguyễn Lê Phước Thịnh
3 tháng 3 2021 lúc 13:48

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

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

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

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

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

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

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