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a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
a/ \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}=2\sqrt{4.2.5\sqrt{4.3}}-2\sqrt{\sqrt{25.3}}-3\sqrt{5\sqrt{16.3}}\)
= \(2.2\sqrt{2.5.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}=4.2\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3.2\sqrt{5\sqrt{3}}\)
= \(\sqrt{5\sqrt{3}}\left(8-2-6\right)=\sqrt{5\sqrt{3}}.0=0\)
b/ \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}=2\sqrt{2.4\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{4.5\sqrt{3}}\)
= \(4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
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) \(\frac{-3\sqrt{3}+3}{2\sqrt{3}-2}=\frac{-3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)}=-\frac{3}{2}\)
\(B=\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}=\frac{9\sqrt{5}+9\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{9\left(\sqrt{5}+\sqrt{3}\right)}{\sqrt{5}+\sqrt{3}}=9\)
\(C=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}.\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
mik chỉnh lại đề
\(D=\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}=\frac{6\sqrt{2}-4\sqrt{3}+2\sqrt{5}}{9\sqrt{2}-6\sqrt{3}+3\sqrt{5}}\)
\(=\frac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}=\frac{2}{3}\)