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11 tháng 2 2022
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{-4}{35}+\dfrac{5}{7}\right)+\dfrac{1}{41}=1+1+\dfrac{1}{41}=\dfrac{83}{41}\)
NP
0
BT
1
28 tháng 1
\(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}-\dfrac{3}{35}+\dfrac{1}{3}-\dfrac{-1}{41}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(-\dfrac{1}{5}-\dfrac{5}{7}-\dfrac{3}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{3+2+1}{6}+\dfrac{-7-25-3}{35}+\dfrac{1}{41}\)
\(=\dfrac{6}{6}+\dfrac{-35}{35}+\dfrac{1}{41}=\dfrac{1}{41}\)
6 tháng 3 2017
A=\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{57}\)+\(\dfrac{1}{3}\)+\(\dfrac{-1}{36}\)
A=(\(\dfrac{-2}{9}\)+\(\dfrac{-3}{4}\)+\(\dfrac{-1}{36}\))+(\(\dfrac{3}{5}\)+\(\dfrac{1}{15}\)+\(\dfrac{1}{3}\))
A=-1+1=0 B=\(\dfrac{1}{2}\)+\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{1}{6}\)+\(\dfrac{-3}{35}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{41}\) B=(\(\dfrac{-1}{5}\)+\(\dfrac{-5}{7}\)+\(\dfrac{-3}{35}\))+(\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{3}\))+\(\dfrac{1}{41}\) B=-1+1+\(\dfrac{1}{41}\)=\(\dfrac{1}{41}\)
4 tháng 5 2020
câu 1 : A=-2/9+-3/4+3/5+1/15+1/57+1/3+-1/36
=(-2/9+-3/4+-1/36)+(3/5+1/15+1/3)
Vậy p/s 1/57 đâu bạn ?
7 tháng 7 2017
-\(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=-\frac{4}{6}+\frac{1}{6}+\frac{3}{4}-\frac{2}{5}=-\frac{2}{4}+\frac{3}{4}-\frac{2}{5}=\frac{1}{4}-\frac{2}{5}=-\frac{3}{20}\)
= \(-\frac{3}{20}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 8 2023
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
\(\dfrac{1}{2}-\left(-\dfrac{2}{5}\right)+\dfrac{1}{3}+\dfrac{5}{7}-\left(-\dfrac{1}{6}\right)+\left(-\dfrac{4}{35}\right)+\dfrac{1}{41}\)
\(=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{1}{3}+\dfrac{5}{7}+\dfrac{1}{6}+-\dfrac{4}{35}+\dfrac{1}{41}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+-\dfrac{4}{35}+\dfrac{5}{7}\right)+\dfrac{1}{41}+\dfrac{1}{3}\)
\(=\dfrac{2}{3}+1+\dfrac{1}{3}+\dfrac{1}{41}\)
\(=\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+1+\dfrac{1}{41}\)
\(=1+1+\dfrac{1}{41}\)
\(=2+\dfrac{1}{41}\)
\(=\dfrac{83}{41}\)
\(#GP\)