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Lời giải:
Xét tử thức:
\(\frac{x-\sqrt{x}}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}}=\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}-1}-\frac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}=\sqrt{x}-\frac{1}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
\(\Rightarrow C=\frac{x-1}{\sqrt{x}}: \frac{\sqrt{x}+1}{x}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{x}{\sqrt{x}+1}=\sqrt{x}(\sqrt{x}-1)\)
\( \left( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } - \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} } \right) \left( \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +1 }{ \sqrt{ x \phantom{\tiny{!}}} -2 } - \dfrac{ \sqrt{ x \phantom{\tiny{!}}} +2 }{ \sqrt{ x \phantom{\tiny{!}}} -1 } \right) \)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(\dfrac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
\( \dfrac{ 1 }{ \sqrt{ x \phantom{\tiny{!}}} -3 } + \dfrac{ 4 }{ \sqrt{ x \phantom{\tiny{!}}} +3 } - \dfrac{ 9- \sqrt{ x \phantom{\tiny{!}}} }{ x-9 } \)(ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\))
\(=\dfrac{1}{\sqrt{x}-3}+\dfrac{4}{\sqrt{x}+3}+\dfrac{\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+3+4\left(\sqrt{x}-3\right)+\sqrt{x}-9}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}-6+4\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{6\sqrt{x}-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=6\cdot\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{6}{\sqrt{x}+3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1+1-\sqrt[]{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{x}{1+\sqrt[]{1-x}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[3]{\left(x+1\right)^3}+\sqrt[3]{x+1}+1}+\dfrac{1}{1+\sqrt[]{1-x}}\right)=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
Ta có: \(\sqrt{48-2\sqrt{135}}-\sqrt{45}+\sqrt{18}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot\sqrt{3}+3}-\sqrt{45}+\sqrt{18}\)
\(=\sqrt{\left(\sqrt{45}-\sqrt{3}\right)^2}-\sqrt{45}+\sqrt{18}\)
\(=\sqrt{45}-\sqrt{3}-\sqrt{45}+\sqrt{18}\)
\(=\sqrt{18}-\sqrt{3}\)
ở VP "+4" nằm ở ngoài căn,đau bụng nên viết vội còn chạy ra WC :P
