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\(x^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3x\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)}\)
\(=2+3x\sqrt[3]{1-\frac{84}{81}}\)
\(=2+3x\sqrt[3]{-\frac{1}{27}}\)
\(=2-x\)
\(\Rightarrow x^3+x-2=0\)
\(\Leftrightarrow x=1\)
\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{8-2\sqrt{15}+8+2\sqrt{15}}{2}-\frac{6+2\sqrt{5}}{4}=\frac{32-6-2\sqrt{5}}{4}=\frac{26-2\sqrt{5}}{4}=\frac{14-\sqrt{5}}{2}\) \(\left(\frac{9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2-\left(\frac{9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)^2=\left(\frac{9-2\sqrt{14}-9-2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)\left(\frac{9-2\sqrt{14}+9+2\sqrt{14}}{\sqrt{7}-\sqrt{2}}\right)=\frac{-72\sqrt{14}}{\sqrt{7}-\sqrt{2}}\)
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)
=>A^3-3A-18=0
=>A=3
b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)
=>B^3-3B-14=0
=>B=2,82
c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)
=>C^3+6C-40=0
=>C=2,84
b. ĐK \(\hept{\begin{cases}x-2\ge0\\y+2014\ge0\\z-2015\ge o\end{cases}\Rightarrow\hept{\begin{cases}x\ge2\\y\ge-2014\\z\ge2015\end{cases}}}\)
Ta có \(\sqrt{x-2}+\sqrt{y+2014}+\sqrt{z-2015}=\frac{1}{2}\left(x+y+z\right)\)
Đặt \(\hept{\begin{cases}\sqrt{x-2}=a\ge0\\\sqrt{y+2014}=b\ge0\\\sqrt{z-2015}=c\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x-2=a^2\\y+2014=b^2\\z-2015=c^2\end{cases}\Rightarrow x+y+z}=a^2+b^2+c^2+3\)
Pt \(\Leftrightarrow a+b+c=\frac{1}{2}\left(a^2+b^2+c^2+3\right)\Leftrightarrow a^2+b^2+c^2+3=2a+2b+2c\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\)\(\Leftrightarrow a=b=c=1\)
\(\Rightarrow\hept{\begin{cases}x-2=1\\y+2014=1\\z-2015=1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2013\\z=2016\end{cases}\left(tm\right)}}\)
Vậy \(x=3;y=-2013;z=2016\)
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))