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a: \(=\dfrac{11}{125}-\dfrac{17}{18}+\dfrac{8}{18}-\dfrac{10}{14}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{11}{125}\)
b: \(=\left(1+2+3+4-1\right)+\left(-\dfrac{1}{2}-\dfrac{2}{3}\right)=9-\dfrac{5}{6}=\dfrac{49}{6}\)
c: \(A=26:\left(6+\dfrac{1}{2}\right)=26:\dfrac{13}{2}=4\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(a.0,\left(13\right)+1,\left(86\right)-\frac{5}{7} \)
\(=86-\frac{5}{7}\)
\(=\frac{602}{7}-\frac{5}{7}\)
\(=\frac{597}{7}\)
\(b.\left[0,\left(4\right)\right]^2-\frac{1}{81}+\frac{22}{27}\)
\(=0-\frac{1}{81}+\frac{66}{81}\)(ở đây quy đồng mẫu số )
\(=-\frac{1}{81}-\frac{66}{81}\)
\(=-\frac{67}{81}\)
Học tốt!!
#Minkk!