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12 tháng 7 2017

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

NV
2 tháng 8 2020

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

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

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

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

\(\Rightarrow A=...\)

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:

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

a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)

b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

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

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

NV
20 tháng 9 2019

ĐKXĐ: ...

\(P=\left(\frac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(P=\left(\frac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\)

\(x=5+\sqrt{2}-4-\sqrt{2}=1\)

\(\Rightarrow P=\frac{1+1}{1+3}=\frac{1}{2}\)

\(P=\frac{\sqrt{x}+1}{\sqrt{x}+3}=1-\frac{2}{\sqrt{x}+3}\)

Do \(\sqrt{x}>0\Rightarrow\sqrt{x}+3>3\Rightarrow\frac{2}{\sqrt{x}+3}< \frac{2}{3}\)

\(\Rightarrow P>1-\frac{2}{3}=\frac{1}{3}\) (đpcm)

20 tháng 9 2019

\(P=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

ĐKXĐ:\(x\ge0;x\ne9\)

\(=\left(\frac{x+3}{x-9}+\frac{1\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x+3}\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\left(\frac{x+3+\sqrt{x}-3}{x-9}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{x+\sqrt{x}}{x-9}.\frac{\sqrt{x-3}}{\sqrt{x}}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b)

\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{5^2+2.5\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{4^2+2.4\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

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

\(=1\)

Thay x=1 vào P ta có:

\(P=\frac{\sqrt{1}+1}{\sqrt{1}-3}\)

\(=\frac{2}{-2}=-1\)

20 tháng 9 2019

huhu cảm ơn cậu nhiều lắm

2 tháng 6 2017
  1. \(\frac{\sqrt{27\left(1-\sqrt{3}\right)^4}}{3\sqrt{15}}=\frac{\sqrt{3.3^2\left(1-\sqrt{3}\right)^4}}{3\sqrt{15}}=\frac{3\left(1-\sqrt{3}\right)^2}{3\sqrt{15}}=\frac{1-2\sqrt{3}+3}{\sqrt{15}}=\frac{4-2\sqrt{3}}{\sqrt{15}}\)
  2. \(=\frac{\sqrt{10}\left(12-8\sqrt{2}+7.15\sqrt{2}\right)}{\sqrt{10}}=12+97\sqrt{2}\)
  3. \(=\sqrt{\frac{x.x\sqrt{y}}{y}}=\sqrt{\frac{x^2}{\sqrt{y}}}=\frac{|x|}{\sqrt[4]{y}}\)