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\(\dfrac{1+3\sqrt{2}-2\sqrt{3}}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}{\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}\)
Tử:
\(\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)
\(=\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)+6\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{6}\) (nhân phân phối)
\(=5\sqrt{3}-7\sqrt{2}\)
Mẫu:
\(\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)
\(=6-\left(5+2\sqrt{6}\right)\)
\(=1-2\sqrt{6}\)
Ta có:
\(\dfrac{5\sqrt{3}-7\sqrt{2}}{1-2\sqrt{6}}\)
\(=\dfrac{\left(5\sqrt{3}-7\sqrt{2}\right)\left(1+2\sqrt{6}\right)}{1-24}\)
\(=\dfrac{5\sqrt{3}+30\sqrt{2}-7\sqrt{2}-28\sqrt{3}}{-23}\)
\(=\dfrac{-23\left(\sqrt{3}-\sqrt{2}\right)}{-23}\)
\(=\sqrt{3}-\sqrt{2}\)
trục căn thức ở mẫu của phân thức
\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
= \(\sqrt{2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
= \(\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}\)
= \(\sqrt{2}+\sqrt{3}+\sqrt{5}\)
thì e chúc sau
a) \(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\dfrac{\left(5+\sqrt{5}\right)\left(5+\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}+\dfrac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\\ =\dfrac{25+5\sqrt{5}+5\sqrt{5}+5+25-5\sqrt{5}-5\sqrt{5}+5}{25-5}\\ =\dfrac{25+25+10}{20}=3\)
b)\(\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+\dfrac{3+\sqrt{2}}{9-2}\\ =\dfrac{7\sqrt{3}-7\sqrt{2}+3+\sqrt{2}}{7}\\ =\dfrac{7\sqrt{3}-6\sqrt{2}+3}{7}\)
please help me
Câu 1:
a: \(=\sqrt{2a\cdot\dfrac{\left(2-a\right)^2}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\dfrac{x\left(x-5\right)^2}{\left(5-x\right)\left(5+x\right)}}=\sqrt{\dfrac{x\left(5-x\right)}{5+x}}\)
c: \(=\sqrt{\dfrac{3a\left(a-b\right)^2}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)