Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B=\(6\frac{4}{9}-4\frac{4}{9}+3\frac{7}{11}\)
B=\(2+3\frac{7}{11}\)
B=\(5\frac{7}{11}\)
B = \(5\frac{7}{11}=\frac{62}{11}\)
C = 1
D = \(\frac{5}{2}=2\frac{1}{2}\)
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\frac{1}{2}.\left(\frac{20}{15}+\frac{6}{15}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{48}{9}\right)\)
\(=\frac{1}{2}.\frac{26}{15}-\frac{3}{4}.\frac{56}{9}\)
\(=\frac{13}{15}-\frac{14}{3}\)
\(=-\frac{19}{5}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\left(\frac{1}{2}.\frac{4}{3}+\frac{1}{2}.\frac{2}{5}\right)-\left(\frac{3}{4}.\frac{8}{9}+\frac{3}{4}.\frac{16}{3}\right)\)
\(=\left(\frac{2}{3}+\frac{1}{5}\right)-\left(\frac{2}{3}+4\right)\)
\(=\frac{2}{3}+\frac{1}{5}-\frac{2}{3}-4\)
\(=\frac{1}{5}-4\)
\(=\frac{1}{5}-\frac{20}{5}=\frac{-19}{5}\)