Bài 1: Tính(hợp lý nếu có thể) a) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\) b) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\) c) \(\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\) d) \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}\)
K
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14 tháng 4 2019
\(A=\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
\(A=\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}-\frac{7}{3}\)
\(A=\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}-\frac{7}{3}\right)+\frac{-5}{4}\)
\(A=2+\left(-1\right)+\frac{-5}{4}\)
\(A=\frac{-1}{4}\)
TY
14 tháng 4 2019
A=\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)\(\frac{7}{3}\)
=\(\frac{-4}{5}+\frac{4}{3}+\frac{-5}{4}+\frac{14}{5}+\frac{-7}{3}\)=\(\left(\frac{-4}{5}+\frac{14}{5}\right)+\left(\frac{4}{3}+\frac{-7}{3}\right)+\frac{-5}{4}\)
=\(\frac{10}{5}+\frac{-3}{3}+\frac{-5}{4}\)=\(2-1+\frac{-5}{4}\)=\(1+\frac{-5}{4}\)=\(\frac{4}{4}+\frac{-5}{4}\)=\(\frac{4-5}{4}\)=\(\frac{-1}{4}\)
27 tháng 7 2019
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
27 tháng 7 2019
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
G
23 tháng 8 2020
a) \(\left|-\frac{6}{7}\right|\div\left(-2\right)^3-\sqrt{\frac{9}{16}}\)
\(=\frac{6}{7}\div\left(-8\right)-\frac{3}{4}\)
\(=\frac{-3}{28}-\frac{3}{4}\)
\(=\frac{-6}{7}\)
b) \(-5\frac{5}{9}\div\left(-1\frac{3}{4}\right)+4\frac{5}{9}\div\left(-1\frac{3}{4}\right)\)
\(=\left(-5\frac{5}{9}+4\frac{5}{9}\right)\div\left(-1\frac{3}{4}\right)\)
\(=\left(-5+\frac{5}{9}+4+\frac{5}{9}\right)\div\frac{-7}{4}\)
\(=\left(-1+\frac{10}{9}\right).\frac{-4}{7}\)
\(=\frac{1}{9}.\frac{-4}{7}\)
\(=\frac{-4}{63}\)
c) \(-63,99-\left(\frac{4}{9}-63,99\right)-\left(-1\frac{2}{3}\right)^2\)
\(=-63,99-\frac{4}{9}+63,99-\left(\frac{-5}{3}\right)^2\)
\(=-63,99-\frac{4}{9}+63,99-\frac{25}{9}\)
\(=\left(-63,99+63,99\right)-\left(\frac{4}{9}+\frac{25}{9}\right)\)
\(=-\frac{29}{9}\)
Bài 1:
a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=4\frac{5}{7}-1\frac{3}{4}\)
\(=\frac{33}{7}-\frac{7}{4}\)
\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)
b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=4\frac{5}{9}-2\frac{3}{4}\)
\(=\frac{41}{9}-\frac{11}{4}\)
\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)
c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)
\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)
\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)
d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)
\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)
\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)