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31 tháng 10 2021

\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)với x >= 0 ; x khác 1 

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

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

số xấu vậy bạn, bạn kiểm tra lại đề nhé 

31 tháng 10 2021

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

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

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

\(P=\frac{\sqrt{x}-1}{\sqrt{x}}\)

25 tháng 5 2017

A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)

A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)

A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)

A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)

B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)

B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)

Bạn tự làm tiếp nhé, mỏi tay quá!!

25 tháng 5 2017

\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)

\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)

\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)

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

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

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

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

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)

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

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

26 tháng 3 2022

b, Ta có \(P=\frac{x\sqrt{x}-\sqrt{x}}{x}=\frac{x-1}{\sqrt{x}}>0\)

\(\Rightarrow\hept{\begin{cases}x-1\ne0\\x-1>0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x>1\\x\ne0\end{cases}}\)

25 tháng 3 2022

với x>hoặc =0;x khác 1 nha