\(\left|-1\frac{5}{6}\right|-\left|+\frac{3}{18}\right|\cdot\sqrt{81}+\sqrt{\frac{9}{24}}\)
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27 tháng 12 2016
a) \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
\(=\frac{15}{6}-\frac{1}{6}.9+\frac{3}{8}\)
\(=\frac{15}{6}-\frac{9}{6}+\frac{3}{8}\)
\(=1+\frac{3}{8}\)
\(=\frac{11}{8}\)
b) \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=2^2.3=12\)
AT
27 tháng 12 2016
a/ \(\left|\frac{-15}{6}\right|-\left|\frac{3}{18}\right|.\sqrt{81}+\sqrt{\frac{9}{64}}\)
= \(\frac{15}{6}-\frac{3}{18}.9+\frac{8}{8}\)
= \(\frac{15}{6}-\frac{3}{2}+\frac{3}{8}\)
= \(\frac{60-36+9}{24}=\frac{33}{24}=\frac{11}{8}\)
b/ \(\frac{6^{15}.9^{10}}{3^{34}.2^{13}}=\frac{\left(2.3\right)^{15}.\left(3^2\right)^{10}}{3^{34}.2^{13}}\) \(=\frac{2^{15}.3^{15}.3^{20}}{3^{34}.2^{13}}=\frac{2^2.3^{35}}{3^{34}}=\frac{4.3}{1}=12\)
1 tháng 2 2020
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
10 tháng 3 2020
d) \(\left(-45,7\right)+\left[\left(+5,7\right)+\left(+5,75\right)+\left(-0,75\right)\right]\)
\(=\left(-45,7\right)+\left[5,7+5,75-0,75\right]\)
\(=\left(-45,7\right)+5,7+5,75-0,75\)
\(=\left[\left(-45,7+5,7\right)\right]+\left[5,75-0,75\right]\)
\(=-40+5=-35\)
e) \(11,26-5,13:\left(5\frac{5}{18}-1\frac{8}{9}\cdot1,25+1\frac{16}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{17}{9}\cdot\frac{5}{4}+\frac{79}{63}\right)\)
\(=11,26-5,13:\left(\frac{95}{18}-\frac{85}{36}+\frac{79}{63}\right)\)
\(=\frac{563}{50}-\frac{513}{100}:\frac{1051}{252}\)
\(=\frac{563}{50}-\frac{513}{100}\cdot\frac{252}{1051}\)
\(=\frac{563}{50}-\frac{129276}{105100}=\frac{21083}{2102}\)
Số lớn quá!
j) \(\sqrt{8^2+6^2}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{64+36}\cdot\sqrt{16}+\frac{1}{2}\cdot\sqrt{\frac{4}{5}}\)
\(=\sqrt{100}\cdot4+\frac{1}{2}\cdot\frac{2\sqrt{5}}{5}\)
\(=10\cdot4+\frac{\sqrt{5}}{5}=40+\frac{\sqrt{5}}{5}=\frac{200+\sqrt{5}}{5}\)
h) Cái đây mình có làm rồi
10 tháng 7 2019
\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
\(=\sqrt{\frac{3+2\sqrt{3}\sqrt{2}+2}{3-2\sqrt{3}\sqrt{2}+2}}+\sqrt{\frac{3-2\sqrt{3}\sqrt{2}+2}{3+2\sqrt{3}\sqrt{2}+2}}\)
\(=\sqrt{\frac{\left(\sqrt{2}+\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)^2}}+\sqrt{\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}+\sqrt{3}\right)^2}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)}\)\
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)+\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}\right)^2+\left(\sqrt{2}-\sqrt{3}\right)^2}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=\frac{5+2\sqrt{6}+5-2\sqrt{6}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}\)
\(=10\)
10 tháng 7 2019
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(\sqrt{2}+3\right)\)
\(=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-3\)
\(=\sqrt{3}-1\)
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