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\(\left(\dfrac{1}{5}\right)^5\cdot5^5\)
\(=\left(\dfrac{1}{5}\cdot5\right)^5\)
\(=\left(\dfrac{5}{5}\right)^5\)
\(=1^5\)
\(=1\)
`@` `\text {Ans}`
`\downarrow`
`(1/5)^5*5^5`
`= (1/5*5)^5`
`= (5/5)^5`
`= 1^5 = 1`
\(\left(\frac{1}{5}\right)^5\cdot5^2\)
\(=\frac{5^2}{5^5}\)
\(=\frac{1}{5^3}\)
\(=\frac{1}{125}\)
hifi thanks
\(\left(\frac{1}{5}\right)^5\cdot5^2\)
\(=\frac{5^2}{5^5}\)
\(=\frac{1}{5^3}\)
\(=\frac{1}{125}\)
hifi thanks
a) 8.(-5).(-4).2 = 8.20.2 = 8.40 = 320
b) \(1\frac{3}{7}+\left(-\frac{1}{3}+2\frac{4}{7}\right)\)
\(=1\frac{3}{7}-\frac{1}{3}+2\frac{4}{7}\)
\(=\left(1\frac{3}{7}+2\frac{4}{7}\right)-\frac{1}{3}=\left(1+2\right)+\left(\frac{3}{7}+\frac{4}{7}\right)-\frac{1}{3}=4-\frac{1}{3}=\frac{11}{3}\)
c) \(\frac{8}{5}\cdot\frac{-2}{3}+\frac{-5\cdot5}{3\cdot5}\)
\(=\frac{8}{5}\cdot\frac{-2}{3}+\frac{-25}{15}=\frac{-16}{15}+\frac{-25}{15}=\frac{-41}{15}\)
d) \(\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\left(-2\right)^2=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{13}{56}\)
a)\(\left(\frac{1}{5}\right)^5\).\(5^5\)=\(\frac{1}{3125}\).3125=1
\(\Leftrightarrow2^x=\dfrac{2^{30}\cdot5^7+2^{13}\cdot5^{27}}{2^{27}\cdot5^7+2^{10}\cdot5^{27}}\)
\(\Leftrightarrow2^x=\dfrac{2^{13}\cdot5^7\left(2^{17}+5^{20}\right)}{2^{10}\cdot5^7\left(2^{17}+5^{20}\right)}=2^3\)
=>x=3
a/ \(\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
b/ \(\left(0,125\right)^3.512=\left(0,125\right).8^3=\left(0,125.8\right)^3=1^3=1\)
c/ \(\left(0,25\right)^4.1024=\left(0,25^2\right)^2.32^2=\left(\dfrac{1}{6}\right)^2.32^2=\left(\dfrac{1}{6}.32\right)^2=16^2\)
\(a,\left(\dfrac{1}{5}\right)^5.5^5=\left(\dfrac{1}{5}.5\right)^5=1^5=1\)
\(b,\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)
\(c,\left(0,25\right)^4.1024=\left(0,25\right)^4.4^4.4=\left(0,25.4\right)^4.4=1^4.4=1.4=4\)
\(\left(\frac{1}{5}\right)^5.5^5=\left(\frac{1}{5}.5\right)^5=1^5=1\)
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