PT
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
CT
1
4 tháng 2 2022
a: \(=5+\dfrac{1}{5}-\dfrac{2}{9}+\dfrac{2}{23}-\dfrac{3}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=-3+\left(-\dfrac{2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\left(\dfrac{1}{5}+\dfrac{2}{23}-\dfrac{3}{35}-\dfrac{2}{7}\right)\)
\(=-3+\dfrac{-12-45+1}{18}+\dfrac{7-10-3}{35}+\dfrac{2}{23}\)
\(=3-\dfrac{28}{9}+\dfrac{-6}{35}+\dfrac{2}{23}=\dfrac{-1417}{7245}\)
DL
0
27 tháng 7 2022
\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=-5+\left(\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{73}{35}\right)+\left(\dfrac{-2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)
\(=-5+\dfrac{7-10+73}{35}+\dfrac{-4-15+1}{18}+\dfrac{1}{23}\)
\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)
NM
20 tháng 7 2023
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
DT
20 tháng 7 2023
a) -1/24 - [ 1/4 - ( 1/2 - 7/8 )]
= -1/24 - [ 1/4 +3/8 ]
= -1/24 - 5/8
= -2/3.
GW
20 tháng 9 2021
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right).\)
\(=5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(-\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(-\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=-5+\frac{70}{35}+\frac{-18}{18}+\frac{1}{23}\)
\(=-5+2+\left(-1\right)+\frac{1}{23}\)
\(=-4+\frac{1}{23}=-\frac{91}{23}\)
LM
Lê Minh Vũ
CTVHS
VIP
20 tháng 9 2021
Tính nhanh nếu có thể:
\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-2\frac{3}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(\left(5+\frac{1}{5}-\frac{2}{9}\right)-\left(2-\frac{1}{23}-\frac{73}{35}+\frac{5}{6}\right)-\left(8+\frac{2}{7}-\frac{1}{18}\right)\)
\(=\)\(5+\frac{1}{5}-\frac{2}{9}-2+\frac{1}{23}+\frac{73}{35}-\frac{5}{6}-8-\frac{2}{7}+\frac{1}{18}\)
\(=\)\(\left(5-2-8\right)+\left(\frac{1}{5}+\frac{73}{35}-\frac{2}{7}\right)+\left(\frac{2}{9}-\frac{5}{6}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+\left(\frac{7}{35}+\frac{73}{35}-\frac{10}{35}\right)+\left(\frac{4}{18}-\frac{15}{18}+\frac{1}{18}\right)+\frac{1}{23}\)
\(=\)\(\left(-5\right)+2+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(\left(-3\right)+-\frac{5}{9}+\frac{1}{23}\)
\(=\)\(-\frac{727}{207}\)
\(\left(5+\dfrac{1}{5}-\dfrac{2}{9}\right)-\left(2-\dfrac{1}{23}-\dfrac{3}{35}+\dfrac{5}{6}\right)-\left(8+\dfrac{2}{7}-\dfrac{1}{18}\right)\)
\(=\left(\dfrac{225}{45}+\dfrac{9}{45}-\dfrac{10}{45}\right)-\left(\dfrac{9660}{4830}-\dfrac{210}{4830}-\dfrac{414}{4830}+\dfrac{4025}{4830}\right)-\left(\dfrac{1008}{126}+\dfrac{36}{126}-\dfrac{7}{126}\right)\)
\(\dfrac{224}{45}-\dfrac{13061}{4830}-\dfrac{1037}{126}=...\)