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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: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a) \(\sqrt{3+\sqrt{5}}\)\(-\sqrt{3-\sqrt{5}}\)\(=\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)\(=\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|}{\sqrt{2}}\)\(=\)\(\frac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}\)\(=\frac{2}{\sqrt{2}}=\sqrt{2}\)
a) \(=\sqrt{\frac{9}{2}}-\sqrt{16.2}+\sqrt{36.2}-\sqrt{81.2}\)
\(=\frac{3}{2}\sqrt{2}-4\sqrt{2}+6\sqrt{2}-9\sqrt{2}\)
\(=\left(\frac{3}{2}-4+6-9\right)\sqrt{2}=\frac{-11}{2}\sqrt{2}\)
b) \(=\frac{\sqrt{5}+3-\sqrt{5}+3}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}.\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\)
\(=\frac{6}{5-9}.\left(-\sqrt{3}\right)=\frac{3}{2}\sqrt{3}\)
c) \(=\left(\frac{a-1-4\sqrt{a}+\sqrt{a}+1}{a-1}\right):\frac{\sqrt{a}\left(\sqrt{a}-2\right)}{a-1}\)
\(=\frac{a-3\sqrt{a}}{a-1}.\frac{a-1}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}-3}{\sqrt{a}-2}\)
\( a)A = \dfrac{{1 + \sqrt 5 }}{{\sqrt 2 + \sqrt {3 + \sqrt 5 } }} + \dfrac{{1 - \sqrt 5 }}{{\sqrt 2 - \sqrt {3 - \sqrt 5 } }}\\ A = \dfrac{{\sqrt 2 \left( {1 + \sqrt 5 } \right)}}{{\sqrt 2 \left( {\sqrt 2 + \sqrt {3 + \sqrt 5 } } \right)}} + \dfrac{{\sqrt 2 \left( {1 - \sqrt 5 } \right)}}{{\sqrt 2 \left( {\sqrt 2 - \sqrt {3 - \sqrt 5 } } \right)}}\\ A = \dfrac{{\sqrt 2 + \sqrt {10} }}{{3 + \sqrt 5 }} + \dfrac{{\sqrt 2 - \sqrt {10} }}{{1 + \sqrt 5 }}\\ A = \dfrac{{\left( {\sqrt 2 + \sqrt {10} } \right)\left( {1 + \sqrt 5 } \right) + \left( {\sqrt 2 - \sqrt {10} } \right)\left( {3 + \sqrt 5 } \right)}}{{\left( {3 + \sqrt 5 } \right)\left( {1 + \sqrt 5 } \right)}}\\ A = \dfrac{{4\sqrt 2 }}{{8 + 4\sqrt 5 }} = - 2\sqrt 2 + \sqrt {10} \\ b)B = \left( {\dfrac{{1 - a\sqrt a }}{{1 - \sqrt a }} + \sqrt a } \right){\left( {\dfrac{{1 - \sqrt a }}{{1 - a}}} \right)^2}\\ B = \left[ {\dfrac{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a + a} \right)}}{{1 - \sqrt a }} + a} \right]{\left[ {\dfrac{{1 - \sqrt a }}{{\left( {1 - \sqrt a } \right)\left( {1 + \sqrt a } \right)}}} \right]^2}\\ B = {\left( {1 + \sqrt a } \right)^2}.\dfrac{1}{{{{\left( {1 + \sqrt a } \right)}^2}}} = 1 \)
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