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\(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}< =>\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{5}\right)^3\)
=> \(x-\frac{2}{15}=\frac{2}{5}\)
x = \(\frac{8}{15}\)
(x-2/15)^3=8/125
(x-2/15)^3=(2/5)^3
(x-2/15)=2/5
x=2/5+2/15
x=8/15
vậy x=8/15

\(\left(2\sqrt{3}\right)^2-\left(3\sqrt{2}\right)^2+\left(4\sqrt{0,5}\right)^2-\left(\frac{1}{5}\sqrt{125}\right)^2\)
\(=2^2.3-3^2.2+4^2.0,5-5\)
\(=12-18+8-5\)
\(=-3\)

(3\(x\) - \(\frac{2}{15}\))\(^3\) = \(\frac{8}{125}\)
(3\(x\) - \(\frac{2}{15}\))\(^3\) = (\(\frac25\))\(^3\)
3\(x\) - \(\frac{2}{15}\) = \(\frac25\)
3\(x\) = \(\frac25\) + \(\frac{2}{15}\)
3\(x\) = \(\frac{6}{15}\) + \(\frac{2}{15}\)
3\(x\) = \(\frac{8}{15}\)
\(x\) = \(\frac{8}{15}\) : 3
\(x\) = \(\frac{8}{15}\) x \(\frac13\)
\(x=\frac{8}{45}\)
Vậy \(x=\frac{8}{45}\)

a) Ta có: \(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}\)
\(=\frac{15}{40}-\frac{8}{40}+\frac{3}{40}\)
\(=\frac{10}{40}=\frac{1}{4}\)
b) Ta có: \(\frac{21}{4}\cdot\frac{3}{8}+\frac{43}{4}\cdot\frac{3}{8}-4\cdot\frac{1}{2}\)
\(=\frac{3}{8}\left(\frac{21}{4}+\frac{43}{4}\right)-2\)
\(=\frac{3}{8}\cdot16-2\)
\(=6-2=4\)
c) Ta có: \(\frac{-5}{9}+\frac{7}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{8}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=\frac{-2}{11}\)
d) Ta có: \(125\%\cdot\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1.5\right)+2016^0\)
\(=\frac{5}{4}\cdot\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}\cdot3+1\)
\(=\frac{15}{16}+\frac{16}{16}=\frac{31}{16}\)

\(\frac{3}{5}-\frac{3}{25}-\frac{3}{125}\)
\(=\frac{75}{125}-\frac{15}{125}-\frac{3}{125}\)
\(=\frac{57}{125}\)
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\(\frac{7}{5}-\frac{7}{25}-\frac{7}{125}\)
\(=\frac{175}{125}-\frac{35}{125}-\frac{7}{125}\)
\(=\frac{133}{125}\)

Câu 1:
a)\(\left(\frac{2}{3}\right)^2=\frac{4}{9}\) b)\(\left(-2\frac{3}{4}\right)^2=\left(-\frac{11}{4}\right)^2=\frac{121}{16}\)
c)\(\left(0,6\right)^4=\left(\frac{3}{5}\right)^4=\frac{81}{625}\) d)\(\left(-\frac{1}{2}\right)^4=\frac{1}{16}\)
e)\(\left(-\frac{1}{5}\right)^5=\frac{-1}{3125}\)
Ta có: \(\left(\frac34x-0,5\right)^3=-\frac{125}{8}\)
=>\(\left(\frac34x-\frac12\right)^3=\left(-\frac52\right)^3\)
=>\(3x-\frac12=-\frac52\)
=>\(3x=-\frac52+\frac12=-\frac42=-2\)
=>\(x=-\frac23\)