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Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
\(M=\left(\frac{a-2\sqrt{a}+1}{a+1}\right):\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)-\left(a+1\right)}\right]\)
\(M=\left[\frac{\left(\sqrt{a}-1\right)^2}{a+1}\right]:\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\left[\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}:\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(M=\frac{\left(\sqrt{a}-1\right)^2}{a+1}.\frac{\left(\sqrt{a}-1\right)\left(a+1\right)}{\left(\sqrt{a}-1\right)^2}=\sqrt{a}+1\)
\(M>1\Leftrightarrow\sqrt{a}-1>1\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>4\)
\(M=\sqrt{3-2\sqrt{2}}-1\)
\(M=\sqrt{\left(\sqrt{2}-1\right)^2}-1=\sqrt{2}-1-1=\sqrt{2}-2\)
\(A=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\left[\left(\frac{1}{2\sqrt{x}}\right)^2-2.\frac{1}{2\sqrt{x}}.\frac{\sqrt{x}}{2}+\left(\frac{\sqrt{x}}{2}\right)^2\right]\)
\(\Leftrightarrow A=\left[\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right]\left(\frac{1}{4x}-\frac{1}{2}+\frac{x}{4}\right)\)
\(\Leftrightarrow A=\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\left(\frac{1}{4x}-\frac{2x}{4x}+\frac{x^2}{4x}\right)\)
\(\Leftrightarrow A=\frac{-4\sqrt{x}}{x-1}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{4\sqrt{x}}{1-x}.\frac{\left(1-x\right)^2}{4x}\)
\(\Leftrightarrow A=\frac{1-x}{\sqrt{x}}\)
b) \(\frac{A}{\sqrt{x}}>1\)
\(\Leftrightarrow\frac{1-x}{\frac{\sqrt{x}}{\sqrt{x}}}>1\)
\(\Leftrightarrow1-x>1\Leftrightarrow x< 0\)
đk: \(a\ge0;a\ne1\)
Ta có:
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-2\left(a^2+2\right)\left(1+\sqrt{a}\right)}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{2a+2\sqrt{a}+2-2a^2\sqrt{a}-2a^2-4-4\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-2a^2\sqrt{a}-2a^2+2a-2\sqrt{a}-2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
\(B=\frac{-a^2\sqrt{a}-a^2+a-\sqrt{a}-1}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)
Tại \(a=\sqrt{2}\) thì giá trị của B là:
\(B=\frac{-\left(\sqrt{2}\right)^2.\left(\sqrt{\sqrt{2}}\right)-\left(\sqrt{2}\right)^2+\sqrt{2}-\sqrt{\sqrt{2}}-1}{\left(1+\sqrt{\sqrt{2}}\right)\left(1-\sqrt{\sqrt{2}}\right)\left(\sqrt{2}+\sqrt{\sqrt{2}}+1\right)}\)
\(B\approx3,45267\)
\(ĐKXĐ:x>1\)
\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{1-a^3}\)
\(=\frac{1-\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{1+\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{\left(1-\sqrt{a}\right)+\left(1+\sqrt{a}\right)}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}\)
\(=\frac{2}{2\left(1-a\right)}+\frac{a^2+2}{a^3-1}=\frac{1}{1-a}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-a^2-a-1+a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\frac{-\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-1}{a^2+a+1}\)
Với \(a=\sqrt{2}\)( thỏa mãn ĐKXĐ ), ta có:
\(B=\frac{-1}{\left(\sqrt{2}\right)^2+\sqrt{2}+1}=\frac{-1}{2+\sqrt{2}+1}=\frac{-1}{3+\sqrt{2}}\)
\(P=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\cdot\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
a) \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}=\left(\sqrt{5}-1\right)-\left(3-\sqrt{5}\right)=2\sqrt{5}-4\)
b) \(\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\) ( \(a\ge0\ne1\))
c) \(\frac{a+\sqrt{a}}{a}=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}.\sqrt{a}}=\frac{\sqrt{a}+1}{\sqrt{a}}=1+\frac{1}{\sqrt{a}}\)(\(a>0\))
d) \(\frac{3+\sqrt{3}}{1+\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=\sqrt{3}\)
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