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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) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
\(A=\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+\frac{5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}\right)\left(15+2\sqrt{6}\right)\)
\(A=\left(\frac{10}{25-24}\right)\left(15+2\sqrt{6}\right)\)
\(A=10\left(15+2\sqrt{6}\right)\)
\(A=150+20\sqrt{6}\)
\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)
\(=14+2\sqrt{49-40}=14+6=20\)
Khi đó:\(A=\sqrt{20}\)
Các câu còn lại bạn làm nốt nhé
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=\frac{\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1}+\frac{\left(2-3\sqrt{2}\right)\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\)
\(=3+\sqrt{2}+\frac{-7\sqrt{2}}{7}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{43+30\sqrt{2}}=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)
Mình đưa ra đáp án thôi nhé :)
a/ \(\left(\sqrt{\frac{5}{3}-\sqrt{\frac{3}{5}}}\right).\sqrt{15}=\sqrt{25-3\sqrt{15}}\)
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}\)
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) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
a) \(A=\frac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\frac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\frac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\)
\(=\frac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{6}\)
\(=\frac{12\sqrt{2}-2\sqrt{18}}{6}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)