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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,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
\(\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\left(1+\dfrac{5}{9}\right)-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\left[\dfrac{448}{891}-\dfrac{25}{99}\right].\dfrac{11}{83}\)
\(=\dfrac{223}{891}.\dfrac{11}{83}\)
\(=\dfrac{223}{6723}\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
Câu 1:
\(\frac{5^4.18^4}{125.9^5.16}\) = \(\frac{5^4.\left(2.9\right)^4}{5^3.9^5.2^4}\) = \(\frac{5^4.2^4.9^4}{5^3.9^5.2^4}\) = \(\frac{5}{9}\)
Câu 2:
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\) = \(\frac{5^{32}.\left(4.5\right)^{43}}{\left(-2.4\right)^{29}.\left(5^3\right)^{25}}\) = \(\frac{5^{32}.4^{43}.5^{43}}{\left(-2\right)^{29}.4^{29}.5^{75}}\) = \(\frac{4^{14}.5^{43}}{\left(-2\right)^{29}.5^{43}}\)
=\(\frac{4^{14}}{\left(-2\right)^{29}}\) = = \(\frac{\left[-2.\left(-2\right)\right]^{14}}{\left(-2\right)^{29}}\) = \(\frac{\left(-2\right)^{14}.\left(-2\right)^{14}}{\left(-2\right)^{29}}\) = \(\frac{\left(-2\right)^{14}}{\left(-2\right)^{15}}\) = \(\frac{-1}{2}\)
H = [0,(32) . 1,(5) - 0,(25)] . \(\frac{11}{83}\)
H = \(\left(\frac{32}{99}.\frac{15}{9}-\frac{25}{99}\right).\frac{11}{83}\)
H = \(\left(\frac{160}{297}-\frac{75}{297}\right).\frac{11}{83}\)
H = \(\frac{85}{297}.\frac{11}{83}\)
H = \(\frac{85}{2241}\)
H = ( 0 . 5 . 0).11 phần 83 (mk ko bt ghi phân số)
H = 0 . 11
H = 0
Study good
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)
=223/6723
ket qua chuan luon do ban k cho minh nha