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20 tháng 9 2021
a) \(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{9-2}=\dfrac{6}{7}\)
b) \(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{5-2\sqrt{15}+3+5+2\sqrt{15}+3}{5-3}=\dfrac{16}{2}=8\)
c) \(=\dfrac{2\left(3\sqrt{2}+4\right)-2\left(3\sqrt{2}-4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{18-16}=\dfrac{16}{2}=8\)
d) \(=\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{\left(2\sqrt{2}-3\sqrt{3}\right)\left(2\sqrt{2}+3\sqrt{3}\right)}=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{8-27}=\dfrac{18\sqrt{3}}{-19}=-\dfrac{18\sqrt{3}}{19}\)
20 tháng 9 2021
\(a,=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\dfrac{6}{3^2-\left(\sqrt{2}\right)^2}=\dfrac{6}{7}\\ b,=\dfrac{2\left(3\sqrt{2}+4\right)-2\left(3\sqrt{2}-4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{18-16}=\dfrac{16}{2}=8\\ c,=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
\(d,=\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)}{8-27}-\dfrac{3}{5\sqrt{3}}=\dfrac{6\sqrt{2}+9\sqrt{3}}{-19}-\dfrac{\sqrt{3}}{5}\\ =\dfrac{-30\sqrt{2}-45\sqrt{3}-19\sqrt{3}}{95}=\dfrac{-30\sqrt{2}-64\sqrt{2}}{95}\)
a: \(A=\left(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right):\dfrac{2}{\sqrt{5}-\sqrt{3}}\)
\(=\left(\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right):\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}\)
\(=\dfrac{\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}=1\)
a:
\(A=\left(\dfrac{\sqrt{14}-\sqrt{7}}{2\sqrt{2}-2}+\dfrac{\sqrt{15}-\sqrt{5}}{2\sqrt{3}-2}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}+\sqrt{5}}{2}\right)\cdot\dfrac{\sqrt{7}-\sqrt{5}}{1}=\dfrac{7-5}{2}=1\)
a:
\(A=\left(a-b\right)\cdot\sqrt{\dfrac{ab}{\left(a-b\right)^2}}\)
\(=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{\left|a-b\right|}\)
a<b<0
=>a-b<0
=>\(A=\left(a-b\right)\cdot\dfrac{\sqrt{ab}}{-\left(a-b\right)}=-\sqrt{ab}\)
a:
\(A=\sqrt{\dfrac{9+12a+4a^2}{b^2}}\)
\(=\sqrt{\dfrac{\left(2a+3\right)^2}{b^2}}=\left|\dfrac{2a+3}{b}\right|\)
a>=-3/2
=>2a+3>=0
b<0
=>\(\dfrac{2a+3}{b}< =0\)
=>\(A=\dfrac{-\left(2a+3\right)}{b}\)
b:
\(A=\left(\dfrac{1}{3-2\sqrt{2}}-\dfrac{6}{2+\sqrt{2}}\right)\left(3+5\sqrt{2}\right)\)
\(=\left(\dfrac{3+2\sqrt{2}}{1}-\dfrac{6\left(2-\sqrt{2}\right)}{2}\right)\left(3+5\sqrt{2}\right)\)
\(=\left(3+2\sqrt{2}-3\left(2-\sqrt{2}\right)\right)\cdot\left(3+5\sqrt{2}\right)\)
\(=\left(5\sqrt{2}-3\right)\left(5\sqrt{2}+3\right)\)
=50-9
=41
b:
\(A=\left(\dfrac{1}{\sqrt{5}-2}-\dfrac{59}{3\sqrt{7}-2}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\dfrac{\sqrt{5}+2}{5-4}-\dfrac{59\left(3\sqrt{7}+2\right)}{63-4}\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}+2-3\sqrt{7}-2\right)\left(3\sqrt{7}+\sqrt{5}\right)\)
\(=\left(\sqrt{5}-3\sqrt{7}\right)\left(\sqrt{5}+3\sqrt{7}\right)\)
=5-63
=-58
b:
\(A=\dfrac{x}{\sqrt{x}-1}+\dfrac{2x-\sqrt{x}}{\sqrt{x}-x}\)
\(=\dfrac{x}{\sqrt{x}-1}-\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\sqrt{x}-1\)