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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 ,...
Ta có :
\(x=\frac{1}{\sqrt{5}-\sqrt{3}}\cdot\sqrt{\frac{10\sqrt{3}-6\sqrt{5}}{5\sqrt{3}+3\sqrt{5}}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{3}}\cdot\sqrt{\frac{2\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{15}\left(\sqrt{5+\sqrt{3}}\right)}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{3}}\cdot\sqrt{\frac{2\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{3}}\cdot\sqrt{\frac{2^2}{\left(\sqrt{5}+\sqrt{3}\right)^2}}\)
\(=\frac{1}{\sqrt{5}-\sqrt{3}}\cdot\frac{2}{\sqrt{5}+\sqrt{3}}\)( Vì \(\sqrt{5}+\sqrt{3}>0\))
\(=\frac{2}{2}=1\)
Thay x= 1 vào A , ta được :
\(A=\left(1^3-1+1\right)^{2019}\)
\(=1\)
Vậy ....
\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\dfrac{5-4}{\sqrt{5}+3-\sqrt{5}}=\dfrac{1}{3}\)A=\(\left(3\left(\dfrac{1}{3}\right)^3+8\left(\dfrac{1}{3}\right)^2+2\right)^{2009}-3^{2009}=3^{2009}-3^{2009}=0\)
a/\(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{x^2}{\sqrt{5}}=\sqrt{20}\Leftrightarrow x^2=\sqrt{100}\Leftrightarrow x=\sqrt{10}\)
b/ \(\sqrt{\left(x-3\right)^2}-9=0\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)
Vậy.......
c/ \(\sqrt{4x^2+4x+1}=6\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\Leftrightarrow\left|2x+1\right|=6\Leftrightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy.......
Ta có :
\(x=\dfrac{\sqrt[3]{10+6\sqrt{3}}\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{\sqrt[3]{3\sqrt{3}+9+3\sqrt{3}+1}\left(\sqrt{3}-1\right)}{\sqrt{5+2\sqrt{5}+1}-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{\sqrt[3]{\left(\sqrt{3}+1\right)^3}\left(\sqrt{3}-1\right)}{\sqrt{\left(\sqrt{5}+1\right)^2-5}}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}\)
\(\Leftrightarrow x=\dfrac{3-1}{1}=2\)
thay x=2 vào biểu thức P ta có :
\(P=\left(2^3-4.2+1\right)^{2015}\)
\(P=1^{2015}=1\)
Nhớ like đúng cho mk nha mọi người
