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a.\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
b. \(\frac{1}{3\sqrt{20}}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
c. \(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{2\left(\sqrt{2}+1\right)}{5\sqrt{2}}=\frac{2\sqrt{2}\left(\sqrt{2}+1\right)}{10}=\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{5}\)
d.\(\frac{\sqrt{21}-\sqrt{7}}{1-\sqrt{3}}=\frac{\sqrt{7}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=\frac{-\sqrt{7}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=-\sqrt{7}\)
e.\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{3-1}=\frac{3\left(\sqrt{3}-1\right)}{2}\)
f.\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\frac{2\left(\sqrt{3}+1\right)}{2}=\sqrt{3}+1\)
a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)
b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)
c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)
d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)
B4
a) \(\frac{9}{\sqrt{3}}=\frac{9\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b)\(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}=\sqrt{5}+\sqrt{2}\)
c)\(\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\frac{\left(\sqrt{2}+1\right)^2}{1}=\left(\sqrt{2}+1\right)^2\)
d)\(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{1}=14\)
B3
a)\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\) \(đk:x\ge1\)
\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\sqrt{x-1}\cdot\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\sqrt{x-1}\cdot\left(-1\right)=-17\)
\(\sqrt{x-1}=17\)
\(\left[{}\begin{matrix}x-1=289\left(tm\right)\\x-1=-289\left(ktm\right)\end{matrix}\right.\)
\(x=290\left(tm\right)\)
Bài 1 :
a, \(\frac{2ab}{\sqrt{a}+\sqrt{b}}=\frac{2ab\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\frac{2ab\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)
b, \(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{\sqrt{10}}{2}\)
c, \(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\sqrt{6}}{2}\)
2. \(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{7}+\sqrt{5}}-\frac{3}{\sqrt{7-2\sqrt{10}}}+\frac{4}{\sqrt{10+2\sqrt{21}}}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2.\left(\sqrt{7}-\sqrt{5}\right)}{7-5}-\frac{3}{\sqrt{2-2\sqrt{10}+5}}+\frac{4}{\sqrt{3+2\sqrt{21}+7}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\frac{2\left(\sqrt{7}-\sqrt{5}\right)}{2}-\frac{3}{\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}}+\frac{4}{\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3}{\left|\sqrt{2}-\sqrt{5}\right|}+\frac{4}{\left|\sqrt{3}+\sqrt{7}\right|}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{3}+\sqrt{7}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3.\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{4.\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{4}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{7}-\sqrt{5}\right)-\left(\sqrt{5}+\sqrt{2}\right)+\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{2}+\sqrt{7}-\sqrt{3}=0\)