tính hợp lí ( nếu có thể)
a)18.17-3*6*7
b)33.(17-5)-7(33-5)
c) (-4-14):(-3)
d)(-25).12(-4).10
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S
10 tháng 3 2023
\(\dfrac{4}{5}\)+\(\dfrac{-5}{4}\)=\(\dfrac{0}{4}\)
\(\dfrac{-1}{3}\)+\(\dfrac{2}{5}-\dfrac{5}{6}\)=\(\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{25}{30}\)=\(\dfrac{-23}{30}\)
\(\dfrac{2}{3}-\dfrac{5}{7}.\dfrac{14}{25}\)=\(\dfrac{2}{3}-\dfrac{2}{5}\)=\(\dfrac{10}{15}-\dfrac{6}{15}\)=\(\dfrac{4}{15}\)
NQ
1
17 tháng 1 2016
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
8 tháng 11 2023
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: \(\dfrac{72}{-6\cdot2+4}=\dfrac{72}{-12+4}=\dfrac{72}{-8}=-9\)
19 tháng 6 2016
a) \(=\left(\left(-\frac{1}{4}-\frac{5}{3}\right)+\frac{7}{33}\right)-\left(-\frac{15}{12}+\frac{6}{11}-\frac{48}{49}\right)\)
\(=\left(-\frac{23}{12}+\frac{7}{33}\right)+\frac{15}{12}-\frac{6}{11}+\frac{48}{49}\)
\(=\left(-\frac{23}{12}+\frac{15}{12}\right)+\left(\frac{9}{33}-\frac{6}{11}\right)+\frac{48}{49}\)
\(=-\frac{2}{3}-\frac{3}{11}+\frac{48}{49}\)
\(=\frac{65}{1617}\)
b) \(=\frac{11}{125}+\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)\)
\(=\frac{11}{125}-\frac{1}{2}+\frac{1}{2}\)
\(=\frac{11}{125}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 2
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.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 5
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
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 5
\(\dfrac{5}{9}.\dfrac{14}{17}+\dfrac{1}{17}.\dfrac{5}{9}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\left(\dfrac{14}{17}+\dfrac{1}{17}\right)+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{5}{9}.\dfrac{15}{17}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{25}{51}+\dfrac{2}{9}+\dfrac{5}{12}\)
=\(\dfrac{691}{612}\)
17 tháng 1 2016
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 ( 17-7)
= 18.10 = 180
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