NQ
3
K
Khách
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7 tháng 7 2017
-\(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=-\frac{4}{6}+\frac{1}{6}+\frac{3}{4}-\frac{2}{5}=-\frac{2}{4}+\frac{3}{4}-\frac{2}{5}=\frac{1}{4}-\frac{2}{5}=-\frac{3}{20}\)
= \(-\frac{3}{20}\)
NH
3
13 tháng 5 2018
\(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
\(=-1+1-\frac{1}{2}=0-\frac{1}{2}\)
\(=\frac{-1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
13 tháng 5 2018
\(\frac{13}{25}+\frac{4}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= \(-1+1-\frac{1}{2}=-\frac{1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
=\(\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
= \(1+1+0,5=2,5\)
LP
1
XO
21 tháng 6 2019
\(\frac{5}{16}+\frac{5}{7}+0,4+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{25}{35}+\frac{2}{5}+\frac{1}{6}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{25}{35}-\frac{4}{35}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{1}{6}+\frac{2}{3}\)
= \(\frac{5}{16}+\frac{2}{5}+\frac{5}{6}\)
= \(\frac{5}{16}+\frac{37}{30}\)
= \(\frac{371}{240}\)
ML
11 tháng 5 2016
\(\frac{-1}{2}\)+0,5+ \(\frac{5}{6}\).(1-\(\frac{5}{4}\))- \(\frac{7}{18}\):\(\frac{-1}{3}\)
=(\(\frac{-1}{2}\)+\(\frac{1}{2}\)) +\(\frac{5}{6}\).(\(\frac{4}{4}\)-\(\frac{5}{4}\))+\(\frac{7}{18}\).\(\frac{3}{1}\)
=0 +\(\frac{5}{6}\).\(\frac{-1}{4}\)+\(\frac{7}{6}\)
=-\(\frac{5}{24}\)+\(\frac{28}{24}\)=\(\frac{23}{24}\)
29 tháng 6 2022
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
11 tháng 3 2019
\(A=\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}\)
\(A=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}\)
\(A=-1+1+\frac{-1}{5}\)
\(A=\frac{-1}{5}\)
11 tháng 3 2019
\(B=\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
\(B=\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-3}{5}+\frac{1}{5}\)
\(B=\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{2}{5}+\frac{-3}{5}+\frac{1}{5}\right)\)
\(B=-1+0\)
\(B=-1\)
Ta có
C=\(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(\left(=\right)C=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{2}{5}+\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(\left(=\right)C=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)+\left(\frac{14}{35}+\frac{25}{35}-\frac{4}{35}\right)+\frac{1}{41}\)
\(\left(=\right)c=1+\left(-1\right)+\frac{1}{41}=\frac{1}{41}\)
vì viết dưới dạng phân số chưa quen nên bn thông cảm cho mk nhé!
C = 0,5 + 1/3 + 0,4 + 5/7 + 1/6 - 4/35 + 1/41
= 5/10 + 1/3 + 4/10 + 5/7 + 1/6 - 4/35 + 1/41
= 1/2 + 1/3 + 2/5 + 5/7 + 1/6 - 4/35 + 1/41
= (1/2 + 1/3 + 1/6) + (2/5 + 5/7 - 4/35) +1/41
= (3/6+ 2/6 + 1/6) + ( 14/35 + 25/35 - 4/35) + 1/41
= 6/6 + 35/35 + 1/41
= 1 + 1 + 1/41 = 2 + 1/41 = 2/1 + 1/41
= 43/41 + 1/41 = 44/41
chúc bn học tốt