Tính:
a) \(\frac{-3}{10}+\frac{-17}{25}\)
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a)\(\frac{{ - 1}}{6} + 0,75 = \frac{{ - 1}}{6} + \frac{3}{4} = \frac{{ - 2}}{{12}} + \frac{9}{{12}} = \frac{7}{{12}}\);
b)\(3\frac{1}{{10}} - \frac{3}{8} = \frac{{31}}{{10}} - \frac{3}{8} = \frac{{124}}{{40}} - \frac{{15}}{{40}} = \frac{{109}}{{40}}\);
c)
\(\begin{array}{l}0,1 + \frac{{ - 9}}{{17}} - \left( { - 0,9} \right) = \frac{1}{{10}} + \frac{{ - 9}}{{17}} + \frac{9}{{10}}\\ = (\frac{1}{{10}} + \frac{9}{{10}}) + \frac{{ - 9}}{{17}} = 1 + \frac{{ - 9}}{{17}} =\frac{{ 17}}{{17}}+\frac{{ - 9}}{{17}}= \frac{8}{{17}}\end{array}\)
a)
\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)
b)
\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)
a) \(log_3\sqrt[3]{3}=\dfrac{1}{2}\)
b) \(log_{\dfrac{1}{2}}8=-3\)
c) \(\left(\dfrac{1}{25}\right)^{log_54}=\dfrac{1}{16}\)
\(\dfrac{10}{17}-\dfrac{5}{13}-\dfrac{-7}{17}-\dfrac{8}{13}+\dfrac{11}{25}\)
\(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}\)
\(=1-1+\dfrac{11}{25}=\dfrac{11}{25}\)
a) bạn nhóm các phan số có mẫu giống nhau nhé
b) mk xin lỗi vì mk ko có thời gian để giải giúp bạn
a,(6/11+16/11)+(7/17+10/17)+(8/25+17/25)=2+1+1=4
2=998+999*1000/999*1000+999-1=1nha tich minh
a) Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)
Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).
b) Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).
c) Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)
Do \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .
d) Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;
\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).
Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
\(\Leftrightarrow A=\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4.4}...\frac{49.51}{50.50}\)
\(\Rightarrow A=\frac{2\cdot4\cdot3\cdot5\cdot...\cdot49.51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(\Rightarrow A=\frac{2\cdot51}{3\cdot50}\)
\(\Rightarrow A=\frac{17}{25}\)
-3/10+-17/25
-15/50+-34/50
-49/50
Bài làm:
\(\frac{-3}{10}+\frac{-17}{25}=\frac{-15}{50}+\frac{-34}{50}=-\frac{49}{50}\)
Sa