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B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
những ai thích xem minecraft và blockman go thì hãy xem kênh youtube của mik kênh mik là M.ichibi các bn nhớ sud và chia sẻ cho nhiều người khác nhé
a) \(\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{ab}-1}\)
b) \(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\sqrt{\frac{x}{y}}\)
c) \(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\frac{3}{7}}\)
d) \(\frac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}=\frac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}\left(\sqrt{1}-\sqrt{2}\right)-\sqrt{3}\left(1-\sqrt{2}\right)}\)
\(=\frac{\left(2\sqrt{5}-\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right)\left(1-\sqrt{2}\right)}=\frac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)
e) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
*****~~~~~~~~~~*****
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+\sqrt{6}}{\sqrt{6}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}\)
\(=\sqrt{3}+\sqrt{6}\)
\(=\sqrt{3}\left(1+\sqrt{2}\right)\)
*****~~~~~~~~~~*****
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=\sqrt{3}+2+\sqrt{2}\)
(Chúc bạn học tốt nha!)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
\(\frac{6-\sqrt{6}}{\sqrt{6}-1}+\frac{6+\sqrt{6}}{\sqrt{6}}\)\(=\frac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\frac{6}{\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{6}}\)\(=\sqrt{6}+\frac{6}{\sqrt{6}}+1\)\(=\sqrt{6}\left(1+\frac{\sqrt{6}}{\sqrt{6}}\right)+1\)\(=\sqrt{6}\left(1+1\right)+1\)\(=\sqrt{6}.2+1\)
\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)\(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-I\sqrt{20}-3I}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)\(=\sqrt{\sqrt{5}-I\sqrt{5}-1I}\)\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)\(=\sqrt{1}=1\)
a)\(\frac{3\sqrt{6}-\sqrt{2}}{1-3\sqrt{3}}=\frac{3\sqrt{3}.\sqrt{2}-\sqrt{2}}{1-3\sqrt{3}}=\frac{\sqrt{2}.\left(3\sqrt{3}-1\right)}{-\left(3\sqrt{3}-1\right)}=-\sqrt{2}\)
b)\(\frac{\sqrt{10}-\sqrt{15}}{\sqrt{8}-\sqrt{12}}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{5}}{2\sqrt{2}-2\sqrt{3}}=\frac{\sqrt{5}.\left(\sqrt{2}-\sqrt{3}\right)}{2.\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{5}}{2}\)
c)\(\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}\)
d)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{5^2.6}-\sqrt{6^2.5}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\sqrt{5}-\sqrt{30}.\sqrt{6}}{\sqrt{5}-\sqrt{6}}=\frac{\sqrt{30}.\left(\sqrt{5}-\sqrt{6}\right)}{\sqrt{5}-\sqrt{6}}=\sqrt{30}\)
e)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{2^2.3}-\sqrt{3^2.2}}{\sqrt{6}}=\frac{\sqrt{6}.\sqrt{2}-\sqrt{6}.\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6}.\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{6}}=\sqrt{2}-\sqrt{3}\)
f)\(\frac{6\sqrt{2}-4}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{16}}{\sqrt{2}}=\frac{6\sqrt{2}-\sqrt{2}.2\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(6-2\sqrt{2}\right)}{\sqrt{2}}=6-2\sqrt{2}\)
g)\(\frac{6-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{36}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.2\sqrt{3}-5\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}.\left(2\sqrt{3}-5\right)}{\sqrt{3}}=2\sqrt{3}-5\)
\(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{7+4\sqrt{3}+7-4\sqrt{3}}{4-3}=14\)
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-\sqrt{2}}+\frac{6}{2+\sqrt{10}}=\frac{\left(\sqrt{15}-\sqrt{12}\right)\left(2+\sqrt{10}\right)+6\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(2+\sqrt{10}\right)}\)
\(=\frac{2\sqrt{15}+5\sqrt{6}-4\sqrt{3}-2\sqrt{30}+6\sqrt{5}-6\sqrt{2}}{5\sqrt{2}-2\sqrt{2}}\)