NT
0
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
NT
2
15 tháng 3 2018
A=-7/21 + (1+1/3)
=-1/3 + 1 + 1/3
=(1/3 + 1/3) +1
=2/3 +1
=2/3 + 3/3
=5/3
B=2/15 + (5/9 + -6/9)
=2/15 + -1/9
=6/45 + 5/45
=11/45
C=(-1/5 + 3/12) + -3/4
=-1/5 + (1/4 + -3/4)
=-1/5 + -1/2
=2/10 + -5/10
=3/10
k cho mik nha
15 tháng 3 2018
A= -7/21 +(1+1/3) = 1
B= 2/15 +(5/9+-6/9) = 61/45C= (-1/5+3/12)+-3/4 = -7/10
14 tháng 7 2019
\(\frac{-7}{21} +\left(1+\frac{1}{3}\right)\)
\(=\frac{-1}{3}+1+\frac{1}{3}=1\)
NY
0
VA
2
18 tháng 3 2020
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-7}{21}+1+\frac{1}{3}\)
\(A=\left(\frac{-7}{21}+\frac{1}{3}\right)+1\)
\(A=0+1\\ A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)
\(B=\frac{2}{15}+\frac{-1}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\left(\frac{3}{12}+\frac{-3}{4}\right)\)
\(C=\frac{-1}{5}+\frac{-1}{2}\)
\(C=\frac{-7}{10}\)
R
18 tháng 3 2020
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-1}{3}+1+\frac{1}{3}\)
\(A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-6}{9}\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-10}{15}\)
\(B=\frac{-8}{15}+\frac{5}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\frac{3}{4}+\frac{-3}{4}\)
\(C=\frac{-1}{5}\)
QT
1
Q
2 tháng 4 2017
Bài 1: Tính
\(\text{1)}\) \(\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{2}.\dfrac{1}{8}\)
\(=\dfrac{5}{8}.\dfrac{7}{30}-\dfrac{5}{8}.\dfrac{1}{2}\)
\(=\dfrac{5}{8}.\left(\dfrac{7}{30}-\dfrac{1}{2}\right)\)
\(=\dfrac{5}{8}.\dfrac{-4}{15}\)
\(=\dfrac{-1}{6}\)
\(\text{2)}\) \(\dfrac{21}{10}.\dfrac{3}{4}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{63}{40}-\dfrac{21}{10}-\dfrac{3}{4}\)
\(=\dfrac{-21}{40}-\dfrac{3}{4}\)
\(=\dfrac{-51}{40}\)
\(\text{3)}\) \(\dfrac{-4}{11}:\dfrac{-6}{11}\)
\(=\dfrac{-4}{11}.\dfrac{11}{-6}\)
\(=\dfrac{4}{6}\)
\(\text{4)}\) \(\dfrac{2}{7}.\dfrac{14}{3}-1\)
\(=\dfrac{4}{3}-1\)
\(=\dfrac{1}{3}\)
\(\text{5)}\) \(\dfrac{4}{7}:\left(\dfrac{1}{5}.\dfrac{4}{7}\right)\)
\(=\dfrac{4}{7}:\dfrac{1}{5}:\dfrac{4}{7}\)
\(=1:\dfrac{1}{5}\)
\(=5\)
\(\text{6)}\) \(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
\(=3+\dfrac{35}{11}.\dfrac{121}{245}\)
\(=3+\dfrac{11}{7}\)
\(=3\dfrac{11}{7}=\dfrac{32}{7}\)
\(\text{7)}\) \(\left(\dfrac{4}{3}+\dfrac{8}{3}\right).\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\left(\dfrac{7}{4}-\dfrac{6}{4}\right):\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\left(\dfrac{6}{5}+\dfrac{12}{5}+\dfrac{1}{5}\right)\)
\(=4.\dfrac{1}{4}:\dfrac{19}{5}\)
\(=1:\dfrac{19}{5}\)
\(=\dfrac{5}{19}\)
\(\text{8)}\) \(\left(\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{\dfrac{1}{9}}{\dfrac{1}{9}}\right):\left(\dfrac{2}{3}+\dfrac{\dfrac{7}{15}}{\dfrac{2}{5}}-\dfrac{1}{6}\right)\)
\(=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{15}:\dfrac{2}{5}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)\)
\(=1:\left(\dfrac{2}{3}+1\right)\)
\(=1:\dfrac{5}{3}\)
\(=\dfrac{3}{5}\)
\(\text{9)}\)
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{75077}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{199}{6613}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\left(\dfrac{7}{1931}-\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\left[\dfrac{3}{50}+\dfrac{9}{2}\right]\)
\(=\dfrac{13235}{13226}:\dfrac{114}{25}\)
\(=\dfrac{330875}{1507764}\)