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\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)
\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)
\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)
a.18.17-3.6.7
=18.17-18.7
=18.(17-7)
=18.10=180
b.54-6.(17+9)
=54-6.26
=54-156=-102
c.33.(17-5)-17.(33-5)
=33.17-33.5-17.33+17.5
=33.5+17.5
=(33+17).5
=50.5=250
Câu 5:
a: \(31\cdot\left(-18\right)+31\cdot\left(-81\right)-31\)
\(=31\left(-18-81-1\right)\)
\(=31\cdot\left(-100\right)=-3100\)
b: \(\left(-12\right)\cdot47+\left(-12\right)\cdot52+\left(-12\right)\)
\(=\left(-12\right)\left(47+52+1\right)\)
\(=-12\cdot100=-1200\)
c: \(13\cdot\left(23+22\right)-3\cdot\left(17+28\right)\)
\(=13\cdot45-3\cdot45\)
\(=45\cdot10=450\)
d: \(-48+48\left(-78\right)+48\left(-21\right)\)
\(=48\left(-1-78-21\right)\)
\(=48\left(-100\right)=-4800\)
Câu 4:
a: \(\left(-6-2\right)\left(-6+2\right)=\left(-8\right)\cdot\left(-4\right)=32\)
b: \(\dfrac{\left(7\cdot3-3\right)}{-6}=\dfrac{21-3}{-6}=\dfrac{18}{-6}=-3\)
c: \(\left(-5+9\right)\cdot\left(-4\right)=4\cdot\left(-4\right)=-16\)
d: \(\dfrac{72}{-6\cdot2+4}=\dfrac{72}{-12+4}=\dfrac{72}{-8}=-9\)
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.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
a)\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}.\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{-3}{5}\)
=\(\dfrac{-2}{5}.1+\dfrac{-3}{5}\)
=\(-\dfrac{2}{5}+\dfrac{-3}{5}\)
=\(-\dfrac{5}{5}\) = -1
\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{691}{612}\)
b) 54-6(17+9)=6.9-6.17-6.9=(6.9-6.9)-17.6=0-102=-102
c) =33.17-33.5-17.33+17.5=33.(17-17)-5.(33-17)=33.0-5.16=0-80=-80
a) \(18.17-3.6.7\)
\(=18.17-18.7\)
\(=18\left(17-7\right)\)
\(=18.10=180\)
c) \(33\left(17-5\right)-17.\left(33-5\right)\)
\(=33.17-33.5-17.33+5.17\)
\(=5\left(17-33\right)\)
\(=5.\left(-16\right)=-80\)
a, 18.17-3.6.7 b,54-6.(17+5) c,33.(17.5)-17.(33-5)
=18.17-18.7 =6.9-6.22 =33.17.5-17.(33-5)
=18.(17-17) =6.(9-22) =17.(33.5)-17.(33-5)
=18.10 =6.(-13) =17.165-17.28
=18.10 =6.(-13) =17.(165-28)
=180 =-78 =17.137
=2329
a) 18.17−3.6.7
= 18.17 - 18 . 7
= 18 ( 17 - 7 )
= 18 . 10 = 180
b) 54−6.(17+9)
= 54 - 102 - 54
= - 102
c) 33.(17−5)−17.(33−5)
= 33. 17 - 33. 5 - 17 .33 + 17 . 5
= 17 . 5 - 33 . 5
= 5 ( 17 - 33 )
= 5. (-16)
= - 80
a) 18.17 - 3.6.7
= 18.17 - 18.7
= 18.(17 - 7 )
= 18 . 10
= 180
b) 54 - 6.(17 + 9)
= 6.9 - 6.17 + 6.9
= (6.9 - 6.9) - 6.17
= 0 - 6.17
= 0 - 102
= -102
c) 33.(17 - 5) - 17.(33 - 5)
= 33.17 - 33.5 - 17 . 33 - 17.5
= 33.(17 - 17) - 5.(33 - 17 )
= 33. 0 - 5.16
= 0 - 80
= -80
a> 18.17-3.6.7
= 18.17-18.7
= 18 ( 17-7)
= 18.10 = 180
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