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a)5.(-8).2.(-3)
=[5.(-2)].[(-8).(-3)]
=-10.24
=-240
b)(45-135+72)-(45+72)
=45-135+72-45-72
=(45-45)+(72-72)-135
=0+0-135
=-135
c)3.(-5)2.2+2.(-5)-20
=3.25.2+2(-5)-20
=75.2+(-10)-20
=150+(-30)
=120
d)86.(-46)+46.27-46.41
=86.(-1).46+46.27-46.41
=-86.46+46.27-46.41
=46(-86+27-41)
=46(-100)
-4600
e)34(15-10)-15(34-10)
=34.15-34.10-15.34+15.10
=(34.15-15.34)+(15.10-34.10)
=[34(15-15)]+[10(15-34)]
=34.0+10(-9)
=0+(-90)
=-90
\(a.36+15+64\\ =\left(36+64\right)+15\\ =100+15\\ =115\\ b.12\cdot36+12\cdot64\\ =12\cdot\left(36+64\right)\\ =12\cdot100\\ =1200\\ c.\left(3^2+2^3\cdot5\right):7\\ =\left(9+8\cdot5\right):7\\ =\left(9+40\right):7\\ =49:7\\ =7\)\(\)
a) \(17\cdot85+15\cdot17-120\)
\(=17\cdot\left(85+15\right)-120\)
\(=17\cdot100-120\)
\(=1700-120\)
\(=1580\)
b) \(27\cdot77+24\cdot27-27\)
\(=27\cdot\left(77+24-1\right)\)
\(=27\cdot100\)
\(=2700\)
a)=(4.25).37
=100.37
=3700
b)=(132+868)+(763+237)+29
=1000+1000+29
=2029
c)=72+[131-1.20]
=49+[131-20]
=49+111
=160
= \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
= \(\frac{2}{7}.2\)
= \(\frac{4}{7}\)
\(\frac{2}{7}.5\frac{1}{4}-\frac{2}{7}.3\frac{1}{4}\)
=> \(\frac{2}{7}.\left(5\frac{1}{4}-3\frac{1}{4}\right)\)
=> \(\frac{2}{7}.2\)
=> \(\frac{4}{7}\)
#Hk_tốt
#Ngọc's_Ken'z
\(a.20-\left[30-\left(5-1\right)^2\right]\) \(b.75-\left(3.5^2-4.2^3\right)\)
\(=20-\left[30-4^2\right]\) \(=75-\left(75-4.2^3\right)\)
\(=20-14=6\) \(=75-\left(75-32\right)\)
\(=75-43=32\)
a: \(=\dfrac{216+3\cdot36+27}{13}=\dfrac{351}{13}=27\)
b: \(=\dfrac{1140-200-40}{9}+3^4=900+81=981\)
a)(-105+107)2 -[(-2)5:(-2)3].(-3)2+20180
= 4-[-32:(-8)].9+1
=4-4.9+1
=4-36+1
=-32 + 1
=-31