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Ta có
\(x=\frac{\sqrt{4+2\sqrt{3}}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}-2}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3}}{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+3.4.\sqrt{5}-8}-2}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)-2}=\frac{1}{5-4-2}=-1\)
Thế vào ta được
\(P=\left(x^2+x+1\right)^{2013}+\left(x^2+x-1\right)^{2013}\)
\(=\left(1-1+1\right)^{2013}+\left(1-1-1\right)^{2013}=1-1=0\)
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
a. ĐK \(x\ge0\)và \(x\ne1\)
A =\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\cdot\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+2\sqrt{x}+1+\sqrt{x}-x-1+\sqrt{x}}\)
\(=\frac{x+1}{4\sqrt{x}}\)
b. Thay \(x=\frac{2-\sqrt{3}}{2}\Rightarrow A=\frac{\frac{2-\sqrt{3}}{2}+1}{4\sqrt{\frac{2-\sqrt{3}}{2}}}=\frac{4-\sqrt{3}}{4\left(\sqrt{3}-1\right)}=\frac{4-\sqrt{3}}{4-4\sqrt{3}}=-\frac{1+3\sqrt{3}}{8}\)
c . Ta có \(A-\frac{1}{2}=\frac{x+1}{4\sqrt{x}}-\frac{1}{2}=\frac{x-2\sqrt{x}+1}{4\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{4\sqrt{x}}>0\)với \(\forall x>0\)và \(x\ne1\)
Vậy A >1/2
- ĐKXĐ : \(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}+1\ne0\\2\sqrt{a}\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}a\ne0\\a\ge0\\a\ne1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
- Ta có phương trình : \(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
=\(\left(\frac{a}{2\sqrt{a}}-\frac{1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2}{a-1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(a-2\sqrt{a}+1-a-2\sqrt{a}-1\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{\sqrt{a}\left(-4\sqrt{a}\right)}{a-1}\right)\)
= \(\left(\frac{a-1}{2\sqrt{a}}\right)\left(\frac{-4a}{a-1}\right)\)= \(\frac{-4a\left(a-1\right)}{2\sqrt{a}\left(a-1\right)}\) = \(\frac{-4a}{2\sqrt{a}}\)
= \(\frac{-4\sqrt{a}\sqrt{a}}{2\sqrt{a}}\) = \(-2\sqrt{a}\)
\(\frac{2+\sqrt{2}}{1+\sqrt{2}}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}+2-\sqrt{2}\)do \(2-\sqrt{2}>0\)
\(=\sqrt{2}+2-\sqrt{2}=2\)