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a) \(\left(\frac{5}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)
\(=\left(\frac{1}{5}-\frac{126}{125}\right):\frac{4}{7}:\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)
\(=\left(\frac{25}{125}-\frac{126}{125}\right):\frac{4}{7}:\left[-\frac{119}{36}\cdot\frac{36}{17}\right]\)
\(=-\frac{101}{125}:\frac{4}{7}:\left(-7\right)=-\frac{101}{125}\cdot\frac{7}{4}\cdot\left(-\frac{1}{7}\right)=\frac{101}{500}\)
b) \(\left(-0,5-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right):\left(-2\right)\)
\(=\left(-\frac{1}{2}-\frac{3}{5}\right):\left(-3\right)+\frac{1}{3}-\left(-\frac{1}{6}\right)\cdot\left(-\frac{1}{2}\right)\)
\(=-\frac{11}{10}:\left(-3\right)+\frac{1}{3}-\frac{1}{12}\)
\(=\frac{11}{30}+\frac{1}{3}-\frac{1}{12}=\frac{37}{60}\)
a) \(\frac{17}{9}-\frac{17}{9}:\left(\frac{7}{3}+\frac{1}{2}\right)\)
= \(\frac{17}{9}-\frac{17}{9}:\frac{17}{6}\)
= \(\frac{17}{9}-\frac{2}{3}\)
= \(\frac{11}{9}\)
b) \(\frac{4}{3}.\frac{2}{5}-\frac{3}{4}.\frac{2}{5}\)
= \(\frac{2}{5}.\left(\frac{4}{3}-\frac{3}{4}\right)\)
= \(\frac{2}{5}.\frac{7}{12}\)
= \(\frac{7}{30}\)
Mình lười làm quá, hay mình nói kết quả cho bn thôi nha
c) -6
d) 3
e) 3
g) 12
h) \(\frac{23}{18}\)
i) \(\frac{-69}{20}\)
k) \(\frac{-1}{2}\)
l) \(\frac{49}{5}\)
Lời giải:
a) $2^3.3^2=8.9=72$
b) $1\frac{4}{5}+\frac{6}{29}-\frac{4}{5}+\frac{23}{29}$
$=1+\frac{4}{5}+\frac{6}{29}-\frac{4}{5}+\frac{23}{29}$
$=1+(\frac{4}{5}-\frac{4}{5})+(\frac{6}{29}+\frac{23}{29})$
$=1+0+\frac{29}{29}=1+0+1=2$
c)
$\frac{-4}{13}.\frac{5}{17}+\frac{-12}{13}.\frac{4}{17}+\frac{4}{13}$
$=\frac{-4}{13}.\frac{5}{17}+\frac{-4}{13}.\frac{12}{17}+\frac{-4}{13}.(-1)$
$=\frac{-4}{13}(\frac{5}{17}+\frac{12}{17}-1)$
$=\frac{-4}{13}.(\frac{17}{17}-1)=\frac{-4}{13}.0=0$
a, \(\left(\frac{1}{5}-\frac{1}{4}\right)^2=\left(\frac{-1}{20}\right)^2=\frac{1}{400}\)
b, \(\left(\frac{5}{3}+\frac{1}{2}\right):\frac{-13}{5}+1\frac{5}{6}\)
=> \(\frac{13}{6}:\frac{-13}{5}+\frac{11}{6}\)
=> \(\frac{13}{6}.\frac{5}{-13}+\frac{11}{6}\)
=> \(\frac{-5}{6}+\frac{11}{6}=\frac{6}{6}=1\)
c, \(\left(\frac{3}{17}\right)^4.\left(\frac{-17}{6}\right)^4\)
=> \(\left(\frac{3}{17}.\frac{-17}{6}\right)^4=\left(\frac{-1}{2}\right)^4=\frac{1}{16}\)
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