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-(0,125)3.803.80=-(0,125.80)3.80=-103.80=-1000.80=-80000
\(\left(0,125\right)^3\times512=\left(\frac{125}{1000}\right)^3\times512=\left(\frac{1}{8}\right)^3\times512=\frac{1^3}{8^3}\times512=\frac{1}{512}\times512=\frac{512}{512}=1\)
\(\dfrac{1}{12345}-3\dfrac{3}{4}+0,125+3\dfrac{3}{4}-\dfrac{1}{8}-\dfrac{1}{12345}\)
\(=\left(\dfrac{1}{12345}-\dfrac{1}{12345}\right)+\left(-3\dfrac{3}{4}+3\dfrac{3}{4}\right)+\left(0,125-\dfrac{1}{8}\right)\)
\(=0+0+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)\)
\(=0\)
#Toru
a, \(\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)=\dfrac{3}{5}.\left(-\dfrac{8}{3}\right)-\dfrac{3}{5}.\left(-\dfrac{2}{3}\right)==\dfrac{3}{5}\left(-\dfrac{8}{3}-\dfrac{2}{3}\right)=\dfrac{3}{5}.\left(-\dfrac{10}{3}\right)=-2\)
b, \(-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)-\left(-\dfrac{21}{15}\right)=-\dfrac{5}{6}.\left(-\dfrac{12}{7}\right)+\dfrac{7}{5}=\dfrac{10}{7}+\dfrac{7}{5}=\dfrac{50+49}{35}=\dfrac{99}{35}\)
a: \(=\dfrac{3}{5}\cdot\left(-\dfrac{8}{3}+\dfrac{-2}{3}\right)=\dfrac{3}{5}\cdot\dfrac{-10}{3}=-2\)
c: \(=\left(0.125\right)^{650}\cdot8^{102}\)
\(=\left(0.125\cdot8\right)^{102}\cdot\left(0.125\right)^{548}\)
\(=\dfrac{1}{8^{548}}\)
a) \(\frac{130^3}{40^3}=\left(\frac{130}{40}\right)^3=\left(\frac{13}{4}\right)^3=\frac{13^3}{4^3}=\frac{2197}{64}\)
b) \(\left(0,125\right)^3.512=\left(\frac{1}{8}\right)^3.2^9=\frac{1^3}{8^3}.2^9=\frac{1}{\left(2^3\right)^3}.2^9=\frac{1}{2^9}.2^9=1\)
c) \(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)^4.2^{10}=\frac{1^4}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=2^2=4\)
a)
\(\begin{array}{l}\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125 \\= \left( {\frac{{ - 3}}{{10}} + \frac{{ - 7}}{{10}}} \right) + \left( {1,125 - 0,125} \right)\\ = - 1 + 1 \\= 0\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9} \\= \frac{8}{3}.\frac{{ - 2}}{{11}} - \frac{8}{3}.\frac{9}{{11}}\\ = \frac{8}{3}.\left( {\frac{{ - 2}}{{11}} - \frac{9}{{11}}} \right)\\ =\frac{{ - 8}}{3}.\frac{-11}{11}\\= \frac{8}{3}.\left( { - 1} \right) \\= \frac{{ - 8}}{3}\end{array}\)
640000