(-1)+(-2)+36+(-17)
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a) \(=\frac{3}{17}-\frac{2}{345}+\frac{5}{12}+\frac{2}{345}-\frac{3}{17}+\frac{1}{12}\)
\(=\left(\frac{3}{17}-\frac{3}{17}\right)+\left(\frac{2}{345}-\frac{2}{345}\right)+\left(\frac{5}{12}+\frac{1}{12}\right)\)
\(=\frac{6}{12}=\frac{1}{2}\)
b) \(=-\frac{49}{25}-\frac{5}{36}+\frac{4}{123}+\frac{49}{25}-\frac{4}{123}-\frac{1}{36}\)
\(=\left(-\frac{49}{25}+\frac{49}{25}\right)+\left(\frac{4}{123}-\frac{4}{123}\right)-\left(\frac{5}{36}+\frac{1}{36}\right)\)
\(=-\frac{6}{36}=-\frac{1}{6}\)
\(\left(\frac{\frac{17}{24}.9\frac{1}{2}-3\frac{1}{4}.\frac{17}{24}}{3\frac{1}{2}.2\frac{13}{36}+2\frac{13}{36}.2\frac{3}{4}}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{\frac{17}{24}.\left(9\frac{1}{2}-3\frac{1}{4}\right)}{2\frac{13}{36}.\left(3\frac{1}{2}+2\frac{3}{4}\right)}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{\frac{17}{24}.\left(\frac{19}{2}-\frac{13}{4}\right)}{\frac{85}{36}.\left(\frac{7}{2}+\frac{11}{4}\right)}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{\frac{17}{24}.\frac{19.2-13}{4}}{\frac{85}{36}.\frac{7.2+11}{4}}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{\frac{17}{24}.\frac{25}{4}}{\frac{85}{36}.\frac{25}{4}}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{17}{24}:\frac{85}{36}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{17}{24}.\frac{36}{85}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{3}{10}-\frac{1}{2}\right)^{-2}\)
\(=\left(\frac{3-5}{10}\right)^{-2}\)
\(=\left(\frac{-1}{5}\right)^{-2}\)
\(=\frac{1}{\left(-\frac{1}{5}\right)^2}=\frac{1}{\frac{\left(-1\right)^2}{5^2}}=\frac{1}{\frac{1}{25}}=25\)
(-1)+(-2) + 36 + (-17) = 36+[(-1) + (-2) + (-17)]= 36 - 20 = 16
a) (-12) + 25 + 75 + 12 = (-12) + 12 + 25 + 75 = 0 +100 = 100.
b) 60 +12 + (-17) + (-43) = 60 + [(-47) + (-43)] +12.
c) (-2) + (-87) + (-18) + 87 = [(-2) + (-18)] + [(-87) + 87].
d) (-1)+(-2) + 36 + (-17) = 36+[(-1) + (-2) + (-17)].
Bài 1:
a: =25+75=100
b: =60-17-43+12=12
c: =-2-18=-20
d: =-3+36-17=36-20=16
Bài 2:
a: =-102
b: =-1000
c: =12x15=180
d: =21x(-10)=-210
\(\left(82-41\cdot2\right):36\cdot\left(32+17+99-81+1\right)\)
\(=\left(82-82\right):36\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0:36\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0\cdot\left[\left(32+17+1\right)+99-81\right]\)
\(=0\)
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\(a,\frac{3}{20}+\frac{3}{20}+\frac{3}{42}+...+\frac{3}{132}\)
\(=3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{11\cdot12}\right)\)
\(=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\right)\)
\(=3\left(\frac{1}{4}-\frac{1}{12}\right)\)
\(=3\cdot\frac{1}{6}=\frac{1}{2}\)
\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)
\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)
\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)
\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)
TL:16