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a) (-2,5. 0,38. 0, 4) - ( 0,125. 3,15. (-8))
=((-2,5.0,4).0,38) - ((-8.0,125).3,15)
= ((-1).0,38) - ((-1).3,15)
= -0,38 - (-3,15)
= 2.77
b) ((-20,83) .0,2 + (-9,17).0,2) : ( 2,47.0,5 - (-3,53).0,5)
= ((-20,83 - 9,17).0,2) : ((2,47 + 3,53).0,5)
= (-6) : 3
= -2
a) (-2,5. 0,38. 0, 4) - ( 0,125. 3,15. (-8))
=((-2,5.0,4).0,38) - ((-8.0,125).3,15)
= ((-1).0,38) - ((-1).3,15)
= -0,38 - (-3,15)
= 2.77
b) ((-20,83) .0,2 + (-9,17).0,2) : ( 2,47.0,5 - (-3,53).0,5)
= ((-20,83 - 9,17).0,2) : ((2,47 + 3,53).0,5)
= (-6) : 3
= -2
a) [(-2,5. 0,38. 0, 4) - ( 0,125. 3,15. (-8)]
=[(-2,5.0,4).0,38] - [(-8.0,125).3,15)]
= [(-1).0,38] - [(-1).3,15]
= -0,38 - (-3,15)
= 2.77
b) [(-20,83) .0,2 + (-9,17).0,2] : [ 2,47.0,5 - (-3,53).0,5]
= [(-20,83 - 9,17).0,2] : [(2,47 + 3,53).0,5]
= (-6) : 3
= -2
a/ \(\left(-2,5.0,38.0,4\right)-\left[0,125.3,15.\left(-8\right)\right]\)
\(=\left(-0,95.0,4\right)-\left[0,39375.\left(-8\right)\right]\)
\(=-0,38-3,15=2,77\)
b/ \(\left[\left(-20,83\right).0,2+\left(-9,17\right).0,2\right]:\left[2,47.0,5-\left(-3,53\right).0,5\right]\)
\(=\left[-4,166+-1,834\right]:\left[1,235--1,765\right]\)
\(=-6:3\)
\(=-2\)
Trả lời
120-(-0,5).(-40).(-5).(-0,2).20.0,25/5+10+10+1995
=120-[(-0,5).(0,2)].[(-40).0,25].[20.(-5)]/2020
=120-0,1.(-10).-100/2020
=120-101/2020
=120-101/2020
=19/2020
\(=\frac{-120+\frac{1}{2}.\left(-40\right).\left(-5\right).\frac{-1}{5}.20.\frac{1}{4}}{5+20.1+1995}\)
\(=\frac{-120+1.\left(-1\right).-5.1.5}{5+1995}\)
\(=\frac{120.-1.1.-5.1.5}{2000}\)
\(=\frac{-120.1\left(-5+5\right)}{2000}\)
\(=0\)
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
\(\left[\left(-20,83\right).0,2+\left(-9,17\right).0,2\right]:\left[2,47.0,5-\left(-3.53\right).0,5\right]\)
\(=\left\{0,2.\left[\left(-20,83\right)+\left(-9,17\right)\right]\right\}:\left[2,47.0,5+3,53.0,5\right]\)
\(=\left[0,2.\left(-30\right)\right]:\left[0,5.\left(2,47+3,53\right)\right]\)
\(=\left(-6\right):\left(0,5.6\right)\)
\(=\left(-6\right):3\)
\(=\left(-2\right)\)
[ (-20,83) . 0,2 + (-9,17) . 0,2 ] : [ 2,47 . 0,5 - (-3.53) . 0,5 ]
= [\(\frac{-2083}{500}\) + \(\frac{-917}{500}\) ] [ \(\frac{247}{200}-\frac{-353}{200}\) ]
= -6 : 3 = -2