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AH
Akai Haruma
Giáo viên
30 tháng 9 2023

Lời giải:
Xét tử thức:

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

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

16 tháng 11 2021

\( \left( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } - \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} } \right) \left( \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ \sqrt{ x \phantom{\tiny{!}}} -2 } - \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +2 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } \right) \)

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

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

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

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

 

\( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -3 } + \dfrac{ 4 }{ \sqrt{ x \phantom{\tiny{!}}} +3 } - \dfrac{ 9- \sqrt{ x \phantom{\tiny{!}}} }{ x-9 } \)(ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\))

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

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

\(=\dfrac{2\sqrt{x}-6+4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{6\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

 

\(A=\frac{x\sqrt{x}-8}{x+2\sqrt{x}+4}+\frac{x\sqrt{x}+27}{x-3\sqrt{x}+9}\) \(=\frac{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}{x+2\sqrt{x}+4}+\frac{\left(\sqrt{x}+3\right)\left(x-3\sqrt{x+9}\right)}{x-3\sqrt{x}+9}\) \(=\sqrt{x}-2+\sqrt{x}+3=2\sqrt{x}+1\)

6 tháng 4 2023

\(\sqrt[ 3 ]{ x-14 \phantom{\tiny{!}}} -\sqrt{ x-1 \phantom{\tiny{!}}} = 3\)

ý bạn là cái này hả :)?

Ta có: \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{45-2\cdot\sqrt{45}\cdot\sqrt{3}+3}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{\left(\sqrt{45}-\sqrt{3}\right)^2}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{45}-\sqrt{3}-\sqrt{45}+\sqrt{18}\)

\(=\sqrt{18}-\sqrt{3}\)

18 tháng 12 2020

ĐK: \(x>1\)

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

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

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

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