Tính giá trị của biểu thức sau:
A= \(-\dfrac{11}{23}.\dfrac{10}{-13}+\dfrac{11}{-13}.\dfrac{3}{-23}-\left(-\dfrac{12}{23}\right)\)
K
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KV
28 tháng 11 2023
a) 7/2 - (3/4 + 1/5)
= 7/2 - 19/20
= 51/20
b) 12/23 . 7/13 + 11/23 . 7/13
= 7/13 . (12/23 + 11/23)
= 7/13 . 1
= 7/13
c) |-2| - (5/9 - 2/3)² : 4/27
= 2 - 1/81 : 4/27
= 2 - 1/12
= 23/12
21 tháng 8 2021
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
7 tháng 3 2022
a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
23 tháng 7 2021
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
23 tháng 7 2021
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
KL
26 tháng 7 2017
a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)
\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)
\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)
\(=\left(-\dfrac{1}{2}\right)2+1\)
\(=-1+1\)
\(=0\)
@Trịnh Thị Thảo Nhi
29 tháng 4 2018
a, 4×(−12)3−2×(−12)2+3×(−12)+14×(−12)3−2×(−12)2+3×(−12)+1
=(−12)[(4×−12)−(2×−12)+3]+1=(−12)[(4×−12)−(2×−12)+3]+1
=(−12)(−2+1+3)+1=(−12)(−2+1+3)+1
=(−12)2+1=(−12)2+1
=−1+1=−1+1
=0=0
23 tháng 3 2022
lần đầu tiên trong đời thấy dấu . là dấu nhân chỉ thấy dấu sao với cả x thôi
T
12 tháng 11 2023
\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)
\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)
\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)
\(=0\)
6 tháng 4 2018
\(A=\dfrac{115-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{403-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{30}}\)
\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)
\(=\dfrac{5}{13}+3\)
\(=\dfrac{5}{13}+\dfrac{39}{13}\)
\(=\dfrac{44}{13}\)
\(=3\dfrac{5}{13}\)
CHÚC BN HC TỐT 
= -11/23.-10/13+-11/23.-3/13-(-12/23)
= -11/23.(-10/13+-3/13)-(-12/23)
= -11/23. -1 -(-12/23)
= 11/23- (-12/23)
= -1/23
Ta có: \(A=\dfrac{-11}{23}\cdot\dfrac{-10}{13}+\dfrac{-11}{13}\cdot\dfrac{-3}{23}-\left(-\dfrac{12}{23}\right)\)
\(=\dfrac{11}{13}\left(\dfrac{10}{23}+\dfrac{3}{23}\right)+\dfrac{12}{23}\)
\(=\dfrac{11}{23}\cdot\dfrac{13}{13}+\dfrac{12}{23}\)
\(=\dfrac{-1}{23}\)