1 4/23+5/11-4/23+6/11+0,5
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NK
30 tháng 12 2022
\(1\dfrac{4}{23}+\dfrac{5}{11}-\dfrac{4}{23}+\dfrac{6}{11}+0,5=\dfrac{27}{23}+\dfrac{5}{11}-\dfrac{4}{23}+\dfrac{6}{11}+0,5\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+0,5=\dfrac{23}{23}+\dfrac{11}{11}+0,5\)
\(=1+1+0,5=2,5\)
30 tháng 12 2022
\(1\dfrac{4}{23}+\dfrac{5}{11}-\dfrac{4}{23}+\dfrac{6}{11}+0,5\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{11}+\dfrac{6}{11}\right)+0,5\)
\(=1+1+0,5=2+0,5=2,5\)
25 tháng 9 2018
1.\(\left(-\frac{6}{5}+\frac{6}{16}-\frac{6}{23}\right):\left(\frac{9}{5}-\frac{9}{16}+\frac{9}{23}\right)\)
\(=6\left(-\frac{1}{5}+\frac{1}{16}-\frac{1}{23}\right):\left(-9\right)\left(\frac{-1}{5}+\frac{1}{16}-\frac{1}{23}\right)\)
\(=6:\left(-9\right)=-\frac{2}{3}\)
2. \(\frac{\frac{3}{7}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{0.5-\frac{1}{3}+\frac{1}{4}}{-\frac{3}{2}+1-\frac{3}{4}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{-3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}-\frac{1}{3}\)
\(=\frac{9}{13}-\frac{5}{15}=\frac{4}{15}\)
24 tháng 10 2021
1: \(\dfrac{4}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=1+\dfrac{1}{2}\)
\(=\dfrac{3}{2}\)
2: \(\left(\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}\right)-\left(\dfrac{79}{67}-\dfrac{28}{41}\right)\)
\(=\dfrac{1}{3}+\dfrac{12}{67}+\dfrac{13}{41}-\dfrac{79}{67}+\dfrac{28}{41}\)
\(=\dfrac{1}{3}\)
TH
1
H
14 tháng 4 2018
Giải:
a) \(25\%-\dfrac{5}{4}+\dfrac{11}{6}\)
\(=\dfrac{1}{4}-\dfrac{5}{4}+\dfrac{11}{6}\)
\(=-1+\dfrac{11}{6}\)
\(=\dfrac{5}{6}\)
Vậy ...
b) \(75\%:\dfrac{11}{5}-\left(0,5\right)^2.\left(-7\right)+2,5\left(\dfrac{23}{3}-\dfrac{17}{3}\right)\)
\(=\dfrac{3}{4}:\dfrac{11}{5}-\dfrac{1}{4}.\left(-7\right)+2,5.2\)
\(=\dfrac{15}{44} +\dfrac{7}{4}+5\)
\(=\dfrac{23}{11}+5\)
\(=\dfrac{78}{11}\)
Vậy ...
c) \(45:\dfrac{18}{7}+50\%-1,25\)
\(=45:\dfrac{18}{7}+\dfrac{1}{2}-\dfrac{5}{4}\)
\(=\dfrac{35}{2}+\dfrac{1}{2}-\dfrac{5}{4}\)
\(=18-\dfrac{5}{4}\)
\(=\dfrac{67}{4}\)
Vậy ...
BA
5
8 tháng 3 2023
\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)
\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)
\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)
\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)
AM
14 tháng 6 2015
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 6
a; \(\dfrac{3}{11}\) + \(\dfrac{5}{-9}\) + \(\dfrac{4}{11}\) - \(\dfrac{4}{9}\) + \(\dfrac{3}{17}\) + \(\dfrac{15}{11}\)
= (\(\dfrac{3}{11}\) + \(\dfrac{4}{11}\) + \(\dfrac{15}{11}\)) - (\(\dfrac{5}{9}\) + \(\dfrac{4}{9}\)) + \(\dfrac{3}{17}\)
= 2 - 1 + \(\dfrac{3}{17}\)
= 1 + \(\dfrac{3}{17}\)
= \(\dfrac{20}{17}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 6
c; N = \(\dfrac{\dfrac{5}{7}-\dfrac{5}{9}-\dfrac{5}{11}}{\dfrac{15}{7}+\dfrac{15}{9}+\dfrac{15}{11}}\)
Phải là - \(\dfrac{5}{7}\) chỗ tử số mới đúng em nhé!
\(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\dfrac{5}{11}+\dfrac{6}{11}+\dfrac{1}{2}=1+1+\dfrac{1}{2}=\dfrac{5}{2}\)