\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{13+30\sqrt{2}+30}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)

\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}+\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}+\sqrt{3}}=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=\frac{5+2\sqrt{6}+\left(5-2\sqrt{6}\right)}{3-2}=10\)

Sai de bai r ban Ngoc oi

Sai chỗ nào bạn Lâm ?

Thích không lập phương thì không lập phương. T dễ tính lắm

\(A=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{40+16\sqrt{13}}+\sqrt[3]{40-16\sqrt{13}}\right)\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{1+3\sqrt{13}+39+13\sqrt{13}}+\sqrt[3]{1-3\sqrt{13}+39-16\sqrt{13}}\right)\)

\(=\dfrac{1}{2}.\left(\sqrt[3]{\left(1+\sqrt{13}\right)^3}+\sqrt[3]{\left(1-\sqrt{13}\right)^3}\right)\)

\(=\dfrac{1}{2}.\left(1+\sqrt{13}+1-\sqrt{13}\right)=\dfrac{2}{2}=1\)

\(A=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\sqrt[3]{5+2\sqrt{13}}=a\)

\(\sqrt[3]{5-2\sqrt{13}}=b\)

\(a^3+b^3=5+2\sqrt{13}+5-2\sqrt{13}=10\)

\(ab=\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}=\sqrt[3]{25-52}=\sqrt[3]{-27}=-3\)

\(A^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(A^3=10-9A\)
\(A^3+9a-10=0\)
\(\left(A-1\right)\left(A^2+A+10\right)=0\)
\(A^2+A+10>0\) mọi A
\(A-1=0\Rightarrow A=1\) là nghiệm duy nhất

KL: A = 1

Ta có  : \(x=\sqrt{\frac{5}{2}}+\sqrt{\frac{2}{5}}=\frac{5+2}{\sqrt{10}}=\frac{7}{\sqrt{10}}>0\)

Do đó : \(A=\sqrt{10x^2}-12x\sqrt{10}+36=x\sqrt{10}-12x\sqrt{10}+36=36-11x\sqrt{10}\)

\(=36-11.\sqrt{10}.\frac{7}{\sqrt{10}}=36-77=-41\)

Đề có sai ko bn , phải là 10x^2 ms khai triển hđt đc chứ

\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-6\sqrt{20}+9}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)

\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)

\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)

\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{5-\sqrt{5}}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}.\)

\(=\left(\left(-\sqrt{7}\right)+\left(-\sqrt{5}\right)\right)\cdot\frac{\sqrt{7}-\sqrt{7}}{1}\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\cdot\frac{\sqrt{7}-\sqrt{5}}{1}\)

\(=\frac{-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{1}\)

\(=\frac{-\left(7-5\right)}{1}=-2\)

a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)

\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)

b: A=2B

=>\(10=4\sqrt{x}-2\)

=>\(4\sqrt{x}=12\)

=>x=9(nhận)

Ta có : \(\sqrt{2008+2\sqrt{2007}}=\sqrt{2007+2\sqrt{2007}+1}=\sqrt{\left(\sqrt{2007}+1\right)^2}=\sqrt{2007}+1\)

\(\sqrt{\left(1-\sqrt{2007}\right)^2}=\sqrt{2007}-1\)

Suy ra \(C=2\sqrt{2007}\)