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\(x=\dfrac{\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}.\left(\sqrt{5}+2\right)=\dfrac{\sqrt[3]{5\sqrt{5}-3.5.2+3\sqrt{5}.4-8}}{\sqrt{5}+\sqrt{9-2.3\sqrt{5}+5}}.\left(\sqrt{5}+2\right)=\dfrac{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}{3}=\dfrac{5-4}{3}=\dfrac{1}{3}\) Thay : \(x=\dfrac{1}{3}\) vào A , ta được :
\(A=\left(\dfrac{3}{27}+\dfrac{8}{9}-\dfrac{3}{3}+1\right)^{2012}=1^{2012}=1\)
Vậy ,...
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
Ta có
\(x=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}-\sqrt{14-6\sqrt{5}}}\)
\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3\cdot5\cdot2+3\sqrt{5}\cdot4-8}}{\sqrt{5}-\sqrt{\left(3-\sqrt{5}\right)^2}}\)
\(=\frac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+3-\sqrt{5}}\)
\(=\frac{\sqrt{5}^2-2^2}{3}=\frac{1}{3}\)
Với \(x=\frac{1}{3}\)thay vào bt ta có
\(A=\left[3\cdot\left(\frac{1}{3}\right)^3+8\cdot\left(\frac{1}{3}\right)^2+2\right]^{2011}\)
\(=3^{2011}\)
\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\dfrac{5-4}{\sqrt{5}+3-\sqrt{5}}=\dfrac{1}{3}\)A=\(\left(3\left(\dfrac{1}{3}\right)^3+8\left(\dfrac{1}{3}\right)^2+2\right)^{2009}-3^{2009}=3^{2009}-3^{2009}=0\)