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a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
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Câu 13 :
\(\left(-\frac{1}{4}+\frac{5}{8}\right)+-\frac{3}{5}\)
\(=\frac{3}{8}-\frac{-3}{5}\)
\(=\frac{39}{40}\)
Câu 14 :
\(M=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{5}{9}.1=\frac{5}{9}\)
Câu 15 :
\(E=\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(E=\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
\(E=0\)
Câu 16 :
\(H=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}+\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{-7}{24}=-3\)
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a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
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a) \(\frac{18}{45}-\frac{4}{12}-\frac{6}{9}-\frac{21}{35}=\frac{2}{5}-\frac{1}{3}-\frac{2}{3}-\frac{3}{5}\)
\(=\left(\frac{2}{5}-\frac{3}{5}\right)-\left(\frac{1}{3}+\frac{2}{3}\right)\)
\(=\frac{-1}{3}-1\)
\(=\frac{-4}{3}\)
b) \(\frac{4}{3}+\frac{3}{5}+\frac{7}{3}+\frac{2}{5}+\frac{1}{3}=\left(\frac{4}{3}+\frac{7}{3}+\frac{1}{3}\right)+\left(\frac{3}{5}+\frac{2}{5}\right)\)
\(=4+1=5\)
c) \(\frac{1}{3}.\frac{4}{5}.\frac{1}{3}.\frac{6}{5}=\frac{8}{75}\)
d) \(\frac{6}{7}.\frac{8}{13}+\frac{6}{7}.\frac{9}{13}-\frac{3}{13}.\frac{6}{7}\)
\(=\frac{6}{7}.\left(\frac{8}{13}+\frac{9}{13}-\frac{3}{13}\right)\)
\(=\frac{6}{7}.\frac{14}{13}\)
\(=\frac{12}{13}\)
1) \(\dfrac{x}{9}=\dfrac{-1}{-4}+\dfrac{1}{12}\)
⇔ \(\dfrac{x}{9}=\dfrac{1}{3}\)
⇔ \(x=\dfrac{9.1}{3}\)
⇔ \(x=3\)
2)\(\dfrac{-1}{3}+\dfrac{1}{4}=\dfrac{-1}{2}x\)
⇔ \(\dfrac{-1}{12}=\dfrac{-1}{2}x\)
⇔\(x=\) \(\dfrac{-1}{12}:\dfrac{-1}{2}\)
⇔\(x=\dfrac{1}{6}\)
3)\(\dfrac{5}{x}+1+\dfrac{4}{x}+1+\dfrac{3}{-13}\)
⇔\(\dfrac{9}{x}+(2+\dfrac{3}{-13})\)
⇔ \(\dfrac{9}{x}+\dfrac{23}{13}\)
1) \(\dfrac{x}{9}=\dfrac{-1}{-4}+\dfrac{1}{12}=\dfrac{1}{3}\) \(\Rightarrow3x=9\) \(\Leftrightarrow x=3\)
2) ĐK: \(x\ne0\)
\(\dfrac{-1}{2x}=-\dfrac{1}{3}+\dfrac{1}{4}=-\dfrac{1}{12}\) \(\Rightarrow2x=12\) \(\Leftrightarrow x=6\)
3) Xem lại đề !!