\(\frac{2}{4}xbằng\frac{7}{12}\)

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27 tháng 3 2017

các bạn có biết ( x + 3 ) + ( x + 5 ) + ( x+ 7 ) + ... + ( x + 45 ) = 1210 ko

nếu biết bảo cho tớ nhé tớ đang cần gấp bài này

5 tháng 4 2016

0.5x - 2/3x = 7/12

<=> 1/2x - 2/3x = 7/12

<=> (1/2 - 2/3).x = 7/12

<=> -1/6.x = 7/12

<=> x = 7/12 : -1/6

<=> x = -7/2

Vậy x = -7/2

25 tháng 3 2017

Chưa ai giải à?

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

\(\Leftrightarrow x-\frac{3}{4}x=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}x=\frac{3}{4}\)

\(\Leftrightarrow x=\frac{3}{4}:\frac{1}{4}=3\)

26 tháng 3 2017

\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)\(=\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{7.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\) 

  \(=\frac{12}{4}:\frac{3}{7}\)

\(=3.\frac{7}{3}=7\)

26 tháng 3 2017

\(\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{3+\frac{3}{13}+\frac{3}{169}+\frac{3}{91}}{7+\frac{7}{13}+\frac{7}{169}+\frac{7}{91}}\)

\(=\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{3\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{9}\right)}{7\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{9}\right)}\)

\(=3:\frac{3}{7}\)

\(=7\)

1 tháng 9 2020

bạn chăm giải thật đó ~!

Bài 1:

a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)

\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)

\(=\frac{-3}{5}\)

b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)

\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)

\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)

c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)

\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)

\(=3+2=5\)

d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)

\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)

\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)

e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)

\(=-1+1+\frac{-7}{9}\)

\(=-\frac{7}{9}\)

f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)

\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)

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

Bài 1:

a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)

\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)

\(=-\frac{3}{7}\)

b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)

\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)

\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)

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

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

\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)

d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)

\(=5^2+2^5-\frac{15^2}{15^2}\)

\(=25+32-1\)

\(=56\)

e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)

\(=\frac{4}{17}+\frac{13}{17}\)

\(=\frac{17}{17}=1\)

g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)

\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)

\(=\frac{7}{12}\cdot4=\frac{7}{3}\)

5 tháng 6 2020

Bài làm

a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)

\(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)

\(-\frac{3}{7}+\frac{7}{2}\)

\(=-\frac{7}{14}+\frac{49}{14}\)

\(=\frac{42}{14}=3\)

b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)

\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)

\(=5-\frac{1}{3}+\frac{2}{9}\)

\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)

\(=\frac{44}{9}\)

c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)

\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)

\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)

\(=\frac{5}{12}+\frac{7}{12}\)

\(=\frac{12}{12}=1\)

d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)

\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)

\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)

\(=-\frac{5}{9}+\frac{1}{90}\)

\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)