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\(a,\left(\frac{3}{15}\right)^{15}:\left(\frac{9}{25}\right)^5=\left(\frac{1}{5}\right)^{15}:\frac{9^5}{25^5}=\frac{1}{5^{15}}.\frac{5^{10}}{9^5}=\frac{1}{5^5.9^5}=\frac{1}{45^5}\)
\(b,5-\left(\frac{-5}{11}\right)^0+\left(\frac{1}{3}\right)^2:3=5-1+\frac{1}{3^2}.\frac{1}{3}=4+\frac{1}{3^3}=\frac{109}{27}\)
\(c,2^3+3\left(\frac{1}{2}\right)^0+\left[\left(-2\right)^2.12\right].8=8+3.1+4.12.8=395\)
a, \(\frac{-1}{39}+\frac{-1}{52}=\frac{-61}{468}\)
b, \(-\frac{9}{6}+-\frac{12}{16}=-\frac{9}{4}\)
\(c,-\frac{2}{5}--\frac{3}{11}=-\frac{7}{55}\)\
\(-\frac{34}{37}\cdot\frac{74}{-58}=\frac{34}{29}\)
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
\(-\frac{2}{6}+\frac{-8}{6}=\frac{-1}{3}+\frac{-4}{3}=\frac{-5}{3}\)
\(\frac{8}{-9}-\frac{9}{-3}=-\frac{8}{9}-\frac{-9}{3}=-\frac{8}{9}-\frac{-27}{9}=\frac{19}{9}\)
\(\frac{2}{5}\cdot\frac{-2}{-5}=\frac{2}{5}\cdot\frac{2}{5}=\frac{4}{25}\)
\(\frac{6}{150}:\frac{6}{-150}=\frac{6}{150}\cdot\frac{-150}{6}=-1\)
d) \(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=-3+\frac{3}{2}+1-5+\frac{6}{5}-\frac{3}{2}\)
\(=-7+\frac{6}{5}=-\frac{29}{5}\)