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\(a\)
\(\sqrt{2,7}\)\(.\)\(\sqrt{1,2}\)
\(=\)\(\sqrt{2,7.1,2}\)
\(=\)\(\sqrt{3,24}\)
\(=\)\(1,8\)
\(b\)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}\)
\(=\)\(\sqrt{85.125.68}\)
\(=\)\(\sqrt{722500}\)
\(=\)\(850\)
học tốt!!!
Ta có: \(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{81}{25}}=\sqrt{\left(\frac{9}{5}\right)^2}=\frac{9}{5}\)
\(\sqrt{2,7}\cdot\sqrt{1,2}\)
\(=\sqrt{2,7\cdot1,2}\)
\(=\sqrt{\frac{27}{10}\cdot\frac{6}{5}}\)
\(=\sqrt{\frac{27}{5}\cdot\frac{3}{5}}\)
\(=\sqrt{\frac{81}{25}}\)
\(=\sqrt{\left(\frac{9}{5}\right)^2}\)
\(=\left|\frac{9}{5}\right|=\frac{9}{5}\)
\(\frac{\sqrt{13,5}}{\sqrt{4,5}}=\sqrt{\frac{13,5}{4,5}}=\sqrt{3}\)
\(\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{3\sqrt{10}+2\sqrt{5}-3\sqrt{6}-2\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(3\sqrt{10}-3\sqrt{6}\right)+\left(2\sqrt{5}-2\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{3\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)+2\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}\)
\(=3\sqrt{2}+2\)
\(=\sqrt{7-2\sqrt{21}+3}+\sqrt{7+2\sqrt{21}+3}\)
\(=\sqrt{\sqrt{7}^2-2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}+\sqrt{\sqrt{7}^2+2\sqrt{7}.\sqrt{3}+\sqrt{3}^2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
\(\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
b) Ta có: \(B=\sqrt{10-2\sqrt{21}}+\sqrt{10+2\sqrt{21}}\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}\)
\(=2\sqrt{7}\)
d) Ta có: \(D=\sqrt{x^2-6x+9}-x\)
\(=\left|x-3\right|-x\)
\(=\left[{}\begin{matrix}x-3-x=-3\left(x\ge3\right)\\3-x-x=-2x+3\left(x< 3\right)\end{matrix}\right.\)
a) \(\sqrt{2,5.2560}=\sqrt{25.256}=\sqrt{25}.\sqrt{256}=5.16=80\)
b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{\frac{7}{2}}.\sqrt{\frac{5}{2}}.\sqrt{7}.\sqrt{\frac{1}{5}}\)
\(=\sqrt{\frac{7}{2}.\frac{5}{2}.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)
c) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1}.\sqrt{0,09}\)
\(=\sqrt{4.121}.\sqrt{9.0,01}=\sqrt{4}.\sqrt{121}.\sqrt{9}.\sqrt{0,01}\)
\(=2.11.3.0,1=6,6\)
Ta có: \(A=\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\)
\(=\sqrt{2011}-\sqrt{2010}< \sqrt{2011}.\sqrt{2010}=B\)
Vậy A<B
\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
\(\sqrt{85}.\sqrt{125}.\sqrt{68}=\sqrt{85.125.68}=\sqrt{5.17.5.25.17.4}\)
\(=\sqrt{5^2.25.17^2.4}=\sqrt{5^2}.\sqrt{25}.\sqrt{17^2}.\sqrt{4}=5.5.17.2=850\)