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\(\left(\frac{2}{7}\right)^{2008}:\left(\frac{4}{49}\right)^{1004}\)
\(=\left(\frac{2}{7}\right)^{2008}:\left[\left(\frac{2}{7}\right)^2\right]^{1004}\)
\(=\left(\frac{2}{7}\right)^{2008}:\left(\frac{2}{7}\right)^{2008}\)
= 1
Học tốt
#Gấu
\(a.9\cdot3^2\cdot\frac{1}{81}=\frac{3^2.3^2.1}{3^4}=\frac{3^4}{3^4}=1\)
\(b.2\frac{1}{2}+\frac{4}{7}:\left(\frac{-8}{9}\right)\)
\(=\frac{5}{2}+\frac{4}{7}.\left(\frac{-9}{8}\right)\)
\(=\frac{5}{2}+\frac{-9}{14}=\frac{13}{7}\)
\(c.3,75.\left(7,2\right)+2,8.\left(3,75\right)\)
\(=3,75.\left(7,2+2,8\right)\)
\(=3,75.10=37,5\)
\(d.\left(\frac{-5}{13}\right).\frac{3}{7}+\left(\frac{-8}{13}\right).\frac{3}{7}+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left[\left(\frac{-5}{13}\right)+\left(\frac{-8}{13}\right)\right]+\left(\frac{-4}{7}\right)\)
\(=\frac{3}{7}.\left(-1\right)+\frac{-4}{7}\)
\(=\frac{-3}{7}+-\frac{4}{7}=-1\)
\(e.\sqrt{81}-\frac{1}{8}.\sqrt{64}+\sqrt{0,04}\)
\(=9-\frac{1}{8}.8+0,2\)
\(=9-1+0,2=8+0,2=8,2\)
a) \(=10\frac{1}{4}\cdot\frac{-5}{3}-8\frac{1}{4}\cdot\frac{-5}{3}-5=\left(10\frac{1}{4}-8\frac{1}{4}\right)\cdot\frac{-5}{3}-5\)
\(=\left(\frac{41}{4}-\frac{33}{4}\right)\cdot\frac{-5}{3}-5=2\cdot\frac{-5}{3}-5\)\(=\frac{-10}{3}-\frac{15}{3}=\frac{-25}{3}\)
b)\(=\frac{5}{7}+1+\frac{2}{7}+\frac{2^{10}\cdot\left(2^3\right)^3}{\left(2^2\right)^9}\)
\(=\frac{5}{7}+\frac{2}{7}+1+\frac{2^{10}\cdot2^9}{2^{27}}\)
\(=1+1+\frac{1}{2^8}=2+\frac{1}{256}=\frac{512}{256}+\frac{1}{256}=\frac{513}{256}\)
\(-\frac{5}{9}\left(\frac{3}{10}-\frac{2}{5}\right)=-\frac{5}{9}\left(\frac{3}{10}-\frac{4}{10}\right)=-\frac{5}{9}.\frac{-1}{10}=\frac{1}{18}\)
\(\frac{1}{2}\sqrt{64}-\sqrt{\frac{9}{25}}+1^{2016}=\frac{1}{2}.8-\frac{3}{5}+1=4+\frac{2}{5}=\frac{22}{5}\)
\(2^8:2^5+3^2.2-12=2^3+9.2-12=8+18-12=8+6=14\)
\(3^x+\sqrt{\frac{16}{81}}-\sqrt{9}+\frac{\sqrt{81}}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+\frac{9}{3}=9\frac{4}{9}\)
\(3^x+\frac{4}{9}-3+3=9\frac{4}{9}\)
\(3^x+\frac{4}{9}=9+\frac{4}{9}\)
\(\Rightarrow3^x=9+\frac{4}{9}-\frac{4}{9}\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
a, \(\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\left(\frac{3}{10}-\frac{4}{10}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{5}{90}\)
\(=\frac{1}{18}\)
b,\(\frac{2}{3}+\frac{-1}{3}+\frac{7}{15}\)
\(=\frac{10}{15}-\frac{5}{15}+\frac{7}{15}\)
\(=\frac{12}{15}\)
\(=\frac{4}{5}\)
c, \(\frac{3}{8}.3\frac{1}{3}\)
\(=\frac{3}{8}.\frac{10}{3}\)
\(=\frac{10}{8}\)
\(=\frac{5}{4}\)
d, \(\frac{-3}{5}+0,8.\left(-7\frac{1}{2}\right)\)
\(=\frac{-3}{5}+\frac{4}{5}.\frac{-15}{2}\)
\(=\frac{-3}{5}+\frac{-60}{10}\)
\(=\frac{-3}{5}+\frac{-30}{5}\)
\(=\frac{-33}{5}\)
e, \(\frac{2}{5}.8\frac{1}{3}+1\frac{2}{3}.\frac{2}{5}\)
\(=\frac{2}{5}.\left(8\frac{1}{3}+1\frac{2}{3}\right)\)
\(=\frac{2}{5}.10\)
\(=4\)
f, \(\frac{3}{7}.19\frac{1}{3}-\frac{3}{7}.33\frac{1}{3}\)
\(=\frac{3}{7}.\left(19\frac{1}{3}-33\frac{1}{3}\right)\)
\(=\frac{3}{7}.-14\)
\(=-6\)
~Study well~
#KSJ
\(\frac{-3}{7}\).\(^{\left(-3\right)^2}\)-\(\sqrt{\frac{4}{49}}\)
= \(\frac{-3}{7}.9-\sqrt{\frac{4}{49}}\)
=\(\frac{-27}{7}-\sqrt{\frac{4}{49}}\)
=\(\frac{-27}{7}-\frac{2}{7}\)
=\(\frac{-29}{7}\)
Chúc bạn học tốt
\(\left|-\frac{3}{7}\right|\cdot(-3)^2-\sqrt{\frac{4}{49}}\)
\(=\frac{3}{7}\cdot9-\frac{2}{7}\)
\(=\frac{27}{7}-\frac{2}{7}=\frac{25}{7}\)