Ta có  

\(\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)^2\)

\(=27+10\sqrt{2}+27-10\sqrt{2}-2\sqrt{\left(27+10\sqrt{2}\right)\left(27-10\sqrt{2}\right)}\)

\(=54-2\sqrt{529}=8\)

\(\Rightarrow\)  \(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}=\sqrt{8}=2\sqrt{2}\)

Xét tử số

\(\left(27+10\sqrt{2}\right)\sqrt{27-10\sqrt{2}}-\left(27-10\sqrt{2}\right)\sqrt{27+10\sqrt{2}}\)

\(=\left(\sqrt{27+10\sqrt{2}}.\sqrt{27-10\sqrt{2}}\right)\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23\left(\sqrt{27+10\sqrt{2}}-\sqrt{27-10\sqrt{2}}\right)\)

\(=23.2\sqrt{2}=46\sqrt{2}\)

Lại có  \(\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2\)

\(=\sqrt{13}-3+\sqrt{13}+3+2\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}+3\right)}\)

\(=2\sqrt{13}+2\sqrt{4}=2\sqrt{13}+4\)

ta bình phương mẫu số

\(\left(\frac{\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}}{\sqrt{\sqrt{13}+2}}\right)^2=\frac{\left(\sqrt{\sqrt{13}-3}+\sqrt{\sqrt{13}+3}\right)^2}{\sqrt{13}+2}\)

\(=\frac{2\sqrt{13}+4}{\sqrt{13}+2}=2\)

Vậy mẫu  \(=\sqrt{2}\)

Vậy  \(x=\frac{46\sqrt{2}}{\sqrt{2}}=46\)  thay vào ta đc A = 92880

\(x=\frac{\left(5+\sqrt{2}\right)^2\sqrt{\left(5-\sqrt{2}\right)^2}-\left(5-\sqrt{2}\right)^2\sqrt{\left(5+\sqrt{2}\right)^2}}{\frac{\sqrt{\left(\sqrt{13}-3\right)\left(\sqrt{13}-2\right)}+\sqrt{\left(\sqrt{13}+3\right)\left(\sqrt{13}-2\right)}}{\sqrt{13-4}}}\)

\(=\frac{\left(5+\sqrt{2}\right)\left(5+\sqrt{2}\right)\left(5-\sqrt{2}\right)-\left(5-\sqrt{2}\right)\left(5-\sqrt{2}\right)\left(5+\sqrt{2}\right)}{\frac{\sqrt{19-5\sqrt{13}}+\sqrt{7+\sqrt{13}}}{3}}\)

\(=\frac{69\left(5+\sqrt{2}-5+\sqrt{2}\right)}{\frac{1}{\sqrt{2}}\left(\sqrt{38-10\sqrt{13}}+\sqrt{14+2\sqrt{13}}\right)}=\frac{276}{\sqrt{\left(5-\sqrt{13}\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}}\)

\(=\frac{276}{5-\sqrt{13}+\sqrt{13}+1}=46\)

\(\Rightarrow A=...\)

a) Ta có: \(P=\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=5+\sqrt{2}-4-\sqrt{2}\)

=1

Thay x=1 vào P, ta được:

\(P=\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

Giúp mình :<

\(\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\sqrt{2}\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3-\sqrt{3-2\sqrt{3}+1}}}=\dfrac{\left(6+2\sqrt{5}\right)\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{\sqrt{2}.\sqrt{4-\sqrt{3}}}=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)

\(=\dfrac{4\sqrt{2}\left(\sqrt{3}+1\right)}{\sqrt{4-\sqrt{3}}}\)

Bài 1:

a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)

\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)

\(=\dfrac{5}{2}\sqrt{5}\)

c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)

\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)

Bài 2:

e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)

Ta có: \(\sqrt{6-x}=3x-4\)

\(\Leftrightarrow6-x=\left(3x-4\right)^2\)

\(\Leftrightarrow9x^2-24x+16+6-x=0\)

\(\Leftrightarrow9x^2-25x+22=0\)

\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)

Vậy: Phương trình vô nghiệm