\(\frac{5}{12}.\frac{-3}{8}-\frac{5}{12}.\frac{5}{12}+2020^0\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(-\frac{5}{12}\right):\frac{7}{3}-\left(-\frac{5}{12}\right):\frac{7}{4}=\left(-\frac{5}{12}\right):\left(\frac{7}{3}-\frac{7}{4}\right)=\left(-\frac{5}{12}\right):\frac{7}{12}=-\frac{5}{7}\)
\(\left[\left(\frac{2}{5}\right)^0\right].\frac{19}{13}-\left(\frac{7}{3}\right)^{2019}.\frac{3}{7}^{2019}\)
\(=\left(\frac{2}{5}\right)^0.\frac{19}{13}-\left(\frac{7}{3}.\frac{3}{7}\right)^{2019}\)
\(=1.\frac{19}{13}-1^{2019}\)
\(=1.\frac{19}{13}-1\)
\(=\frac{19}{13}-1\)
\(=\frac{6}{13}\)
Bài giải
a, \(\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\text{ : }\frac{7}{3}-\left(-\frac{5}{12}\right)\text{ : }\frac{7}{4}\)
\(=\left(-\frac{5}{12}\right)\cdot\frac{3}{7}-\left(-\frac{5}{12}\right)\cdot\frac{4}{7}\)
\(=\frac{-15}{84}+\frac{20}{84}=\frac{5}{84}\)
b, \(\left[\left(\frac{2}{5}\right)^0\right]^{2020}\cdot\frac{19}{37}-\left(\frac{7}{3}\right)^{2019}\cdot\frac{3^{2019}}{7}\)
\(=1^{2020}\cdot\frac{19}{37}-\frac{7^{2019}}{3^{2019}}\cdot\frac{3^{2019}}{7}\)
\(=\frac{19}{37}-7^{2018}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right)\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=-\frac{3}{5}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(\frac{-5}{8}\right)-\left(\frac{-5}{10}\right)+\left(\frac{-5}{11}\right)+\left(\frac{-5}{12}\right)}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right).\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=\frac{-3}{5}\)
a) $\frac{1}{6} \times \frac{2}{3} = \frac{{1 \times 2}}{{6 \times 3}} = \frac{2}{{18}} = \frac{1}{9}$
b) $\frac{6}{5} \times \frac{3}{8} = \frac{{6 \times 3}}{{5 \times 8}} = \frac{{18}}{{40}} = \frac{9}{{20}}$
c) $\frac{4}{3} \times \frac{8}{9} = \frac{{4 \times 8}}{{3 \times 9}} = \frac{{32}}{{27}}$
d) $\frac{5}{{12}} \times \frac{{12}}{5} = \frac{{5 \times 12}}{{12 \times 5}} = \frac{{60}}{{60}} = 1$
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(M=\frac{3}{-5}=\frac{-3}{5}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
a) =-5/7 +7/8-2/7+1/8- -1/12+ -13/12
=(-5/7-2/7)+(7/8+1/8)-(-1/12--13/12)
=-7/7+8/8 - 12/12
= -1+1+1
=1
b)= ( -3/8+11/8)-(12/11+ -1/11)+(-3/5- 2/5)
= 1- 1 + (-1)
=-1
= \(\frac{5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+2\frac{1}{8}\)
=> \(\frac{5}{8}.1+2\frac{1}{8}\)=> \(\frac{5}{8}+\frac{17}{8}=\frac{22}{8}=\frac{11}{4}\)
#Hk_tốt
#kEn'Z
\(\frac{-5}{8}.\frac{5}{12}+\frac{-5}{8}.\frac{7}{12}+2\frac{1}{8}=\frac{-5}{8}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{17}{8}=\frac{-5}{8}.1+\frac{17}{8}=\frac{-5}{8}+\frac{17}{8}=\frac{12}{8}=\frac{3}{4}\)