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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt tổng trên là A
\(A=\left[\left(-37\right).86\right]+\left(37.76\right)\)
Áp dụng qui tắc dấu ngoặc, ta có:
\(A=\left[\left(-37\right)+37\right].\left(86+76\right)\)
\(A=0.\left(86+76\right)\)
\(A=0.162\)
\(A=0\)
Vậy A = 0
\(-37\times86+37\times76\)
\(=37\times\left(-86+76\right)\)
\(=37\times\left(-10\right)\)
\(=-370\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=72\cdot\left(-200\right)=-14400\\ b,=\left(125\cdot8\right)\left[\left(-5\right)\left(-20\right)\right]=1000\cdot100=1000000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(123-123\right)+\left(54+46\right)=100\\ b,=\left(-64+64\right)+\left(71-111\right)=-40\)
![](https://rs.olm.vn/images/avt/0.png?1311)
#\(N\)
`a, -677*54+54.(-125)+2.54`
`=54.(-677+-125+2)`
`=54.(-800)`
`=-43200`
`b, -(324-267+252)+(133-76)`
`=-324+267-252+133-76`
`= (-324+-76)+(133+267)-252`
`=-400+400-252`
`=-252`
a: =54(-677-125+2)
=54*(-800)
=-43200
b: =-324+267-252+133-76
=-400+400-252
=-252
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: =>x-35=-23
=>x=12
b: =>|x-8|=13
=>x-8=13 hoặc x-8=-13
=>x=21 hoặc x=-5
Bài 1:
a: =42-98-42+12-12=-98
b: =10x4x3x(-25)=40x(-25)x3=-1000x3=-3000
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2\cdot53\cdot12+4\cdot6\cdot87-3\cdot8\cdot40\)
\(=24\left(53+87-40\right)\)
=2400
b: \(5\cdot7\cdot77+7\cdot60+49\cdot25-45\cdot42\)
\(=35\cdot77+35\cdot12+49\cdot25-45\cdot42\)
\(=35\cdot89+35\left(7\cdot5-9\cdot6\right)\)
\(=35\cdot70=2450\)
a) 2.53.12+4.6.87-3.8.40
=24.53+24.87-24.40
=24.(53+87-40)
=24.100
=2400
b) 5.7.77-7.60+49.25-45.42
=5.7.77-7.60+7.7.25-45.6.7
=7.(5.77-60+7.25-6.45)
=7.230
=1610
B1:A=2.17.12+4.6.21+8.3.62
A=(2.12).17+(4.6).21+(8.3).62
A=24.17+24.21+24.62
A=24.(17+21+62)
A=24.100
A=2400
B2:
B=37.24+37.76+63.79+63.21
B=37(24+76)+63(79+21)
B=37.100+63.100
B=100.(63+37)
B=100.100
B=10000