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28 tháng 5 2020

1) \(\left(\frac{7}{8}-\frac{3}{4}\right).1\frac{1}{3}-\frac{2}{7}.\left(3,5\right)^2\)

= \(\frac{1}{8}.\frac{4}{3}-\frac{2}{7}.\frac{49}{4}\) = \(\frac{-10}{3}\)

2) \(\frac{5}{4}:\frac{1}{2}-\frac{3}{4}\) = \(\frac{7}{4}\)

5 tháng 4 2019

\(A=\frac{1}{4}:\left(10,3-9,8\right)-\frac{3}{4}\)

\(A=\frac{1}{4}:\left(0,5\right)-\frac{3}{4}\)

\(A=\frac{1}{4}:\frac{1}{2}-\frac{3}{4}\)

\(A=\frac{1}{4}\cdot2-\frac{3}{4}=\frac{1}{2}-\frac{3}{4}=\frac{2}{4}-\frac{3}{4}=\frac{-1}{4}\)

5 tháng 4 2019

Bài 1:

a, \(A=\frac{1}{4}:0,5-\frac{3}{4}\)

       \(=\frac{1}{2}-\frac{3}{4}\)

         \(=\frac{-1}{4}\)

b, \(B=\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.\left(-2\right)^2\)

       \(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}\)

       \(=\frac{6}{7}+\frac{-5}{8}\)

       \(=\frac{13}{56}\)

Hk tốt

31 tháng 5 2017

c) x=-2 nha

d) =\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+......+\(\frac{1}{11.12}\)

=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+.....+\(\frac{1}{11}\)-\(\frac{1}{12}\)

=\(\frac{1}{5}\)-\(\frac{1}{12}\)\(\frac{7}{60}\)

31 tháng 5 2017

bạn ơi kết quả là 7/60

...
Đọc tiếp

\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

TRÌNH BÀY GIÚP MÌNH NHA 

0
30 tháng 4 2019

\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{1}{2}\)

\(=\frac{1}{4}+\frac{2}{4}\)

\(=\frac{3}{4}\)

30 tháng 4 2019

\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)

\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)

\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)

\(=\frac{25}{20}-\frac{-1}{20}\)

\(=\frac{26}{20}\)

\(=\frac{13}{10}\)

16 tháng 7 2018

\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)

\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)

\(A=1-1=0\)

\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)

\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)

\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)

\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)

\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)