11+2 23-12
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DN
1
TT
13 tháng 9 2021
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TG
4 tháng 4 2022
11/23 × 12/17+11/23×5/17+12/23
=(11/23+12/23) x (12/17+5/17)
=1 x 1
=1
VQ
4 tháng 4 2022
11/23 × 12/17+11/23×5/17+12/23
=11/23 × 12/17+11/23×5/17+12/23x1
=11/23 × (12/17+5/17+1)
=11/23 × 2
= 22/23
11 tháng 3 2016
5/11+16/22+(-12/4) +(-2/11)
= -2
7/23+(-10/18+(-4/9)+16/23
= 0
bn quy đồng mẫu số nha
duyệt đi
28 tháng 2 2022
a: =6/35x39/54=13/105
b: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}=\dfrac{-5}{7}+\dfrac{12}{7}=1\)
4 tháng 7 2021
\(a,\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)\)
\(=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{132}.\dfrac{132}{103}=\dfrac{115}{103}\)
\(b,\left(-\dfrac{1}{2}\right)^2-\left(-2\right)^2-5^0\)
\(=\dfrac{1}{4}-4-1=-\dfrac{19}{4}\)
\(c,12\dfrac{1}{3}-\dfrac{5}{7}:\left(24-23\dfrac{5}{7}\right)\)
\(=\dfrac{37}{3}-\dfrac{5}{7}.\dfrac{2}{7}=\dfrac{37}{3}-\dfrac{10}{49}=\dfrac{1783}{147}\)
8 tháng 3 2023
\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)
\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)
\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)
\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)
29 tháng 5 2018
\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
11 + 2 = 13
23 - 12 = 11