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P = \(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)
P = \(\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{x-2\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
P = \(\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
Với \(x=6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
=> P = \(\frac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}-3}=\frac{\sqrt{5}-1+1}{\sqrt{5}-1-3}=\frac{\sqrt{5}}{\sqrt{5}-4}=\frac{\sqrt{5}\left(\sqrt{5}+4\right)}{\left(\sqrt{5}-4\right)\left(\sqrt{5}+4\right)}=\frac{5+4\sqrt{5}}{-11}\)
=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)
Đặt \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}.\)
\(\Rightarrow x^3=\sqrt{5}+2-3\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\left(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\right)-\sqrt{5}+2\)
\(=4-3\sqrt[3]{5-4}.x\)( Vì \(x=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\))
\(=4-3x\)
\(\Rightarrow x^3+3x-4=0\Leftrightarrow\left(x^3-1\right)+\left(3x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+4\right)=0\Leftrightarrow x-1=0\)( Vì \(x^2+x+4=\left(x+\frac{1}{2}\right)^2+\frac{15}{4}>0\))
\(\Leftrightarrow x=1\)hay \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}=1\)
\(a,ĐKXĐ:a\ge0;a\ne4\)
\(P=\frac{\sqrt{a}+1}{\sqrt{a}-2}+\frac{2\sqrt{a}}{\sqrt{a}+2}-\frac{5\sqrt{a}+2}{a-4}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}-2}+\frac{2\sqrt{a}}{\sqrt{a}+2}-\frac{5\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)+2\sqrt{a}\left(\sqrt{a}-2\right)-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\frac{3\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)
\(b,P=2\Rightarrow\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)
\(\Rightarrow3\sqrt{a}=2\left(\sqrt{a}+2\right)\)
\(\Rightarrow3\sqrt{a}=2\sqrt{a}+4\)
\(\Rightarrow3\sqrt{a}-2\sqrt{a}=4\)
\(\Rightarrow\sqrt{a}=4\)
\(\Rightarrow a=16\)
Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
\(\frac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left|\sqrt{5}-1\right|=\sqrt{5}-\sqrt{5}+1=1\)