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Đặt \(\sqrt[4]{5}=x\) thì \(x^4=5\). Ta có :
A = \(\frac{2}{\sqrt{4-3x+2x^2-x^3}}\)= \(\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}\)= \(\frac{2\left(x+1\right)}{\sqrt{-x^5+5x+4}}\)
Ta thấy \(-x^5+5x\) = \(x\left(5-x^4\right)\)= \(0\)
nên A = \(\frac{2\left(x+1\right)}{\sqrt{4}}\)= \(x+1\)=\(\sqrt[4]{5}+1\)
\(\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{9+4\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2^2+2.2\sqrt{5}+\sqrt{5^2}\right)}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=\sqrt[3]{2-\sqrt{5}}\left(\sqrt[6]{\left(2+\sqrt{5}\right)^2}+\sqrt[3]{2+\sqrt{5}}\right)\)
\(=2\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{2+\sqrt{5}}=2\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=2\sqrt[3]{4-5}=2\sqrt[3]{-1}=-1.2=-2\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)
\(=0.\sqrt{5}=0\)
b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)
\(=-\left(-\sqrt{2}+\sqrt{3}\right)=\sqrt{2}-\sqrt{3}\)
a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)
\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)
\(=0.\sqrt{5}\)
\(=0\)
b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)
\(=\frac{\left(5-2\sqrt{6}\right).\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\sqrt{3}-\sqrt{2}}{-1}\)
\(=\sqrt{2}-\sqrt{3}\)
NX \(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}\) =\(\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}-1\right)}{\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}+1\right)^2}\)
=\(\frac{\left(\left(\sqrt{n+1}-\sqrt{n}\right)^2-1^2\right)}{n+1-n-1-2\sqrt{n}}\) \(=\frac{n+1+n-2\sqrt{\left(n+1\right)n}-1}{-2\sqrt{n}}=\frac{2n-2\sqrt{n\left(n+1\right)}}{-2\sqrt{n}}\)
=\(\frac{n-\sqrt{n\left(n+1\right)}}{-\sqrt{n}}=\frac{n}{-\sqrt{n}}+\frac{\sqrt{n\left(n+1\right)}}{\sqrt{n}}=-\sqrt{n}+\sqrt{n+1}\)
thay vao Q ta co
Q= \(-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-...-\sqrt{2012}+\sqrt{2013}=-\sqrt{2}+\sqrt{2013}\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
de thoi
de lam