`3/2-5+1/10=15/10-50/10+1/10=[15-50+1]/10=[-34]/10=-17/5`

106.21=(100+6).21

=100.21+6.21

=2100+6.(20+1)

=2100+120+6

=2226

106.21=(100+6).21

=100.21+6.21

=2100+6.(20+1)

=2100+120+6

=2226

`a// (-2020)+(-598)+(-201)+498+301`

`=-2020-598-201+498+301`

`=-2020-(598-498)-(201-301)`

`=-2020-100-(-100)=-2020-[100+(-100)]=-2020`

`b// (-511)-(-11)`

`=-511+11=-500`

\(a)\left(-2020\right)+\left(-598\right)+\left(-201\right)+498+301\)

\(=\left(-2020\right)+\left\{[\left(-598\right)+498]+[\left(-201\right)+301]\right\}\)

\(=\left(-2020\right)+\left\{\left(-100\right)+100\right\}\)

\(=2020.\)

\(b)\)\(\left(-511\right)-\left(-11\right)\)

\(=\left(-511\right)+11\)

\(=-500\)

-56 : 7 = - 8

-3x = 264

x = -264 : 3

x = -88

100 : (x-7) = 1

x-7 = 100: 1

x - 7 = 100

x = 100 + 7 

x = 107

\(a)\left(-56\right):7=-\left(56:7\right)=-8\)

\(b)\left(-3\right).x=264\)

             \(x=264:\left(-3\right)\)

             \(x=-88\)

\(c)100:\left(x-7\right)=1\)

               \(x-7=100\)

                      \(x=107\)

Thời gian ô tô đi hết quãng đường là:

157,5:52,5=3(h)

Đs:3h.

\([\left(-9\right).\left(-9\right).\left(-9\right)+9^3]:8^{10}\)

\(=[\left(-729\right)+729]:8^{10}\)

\(=0\)

= ( -729 + 729) : 8^10

= 0: 8^10

= 0

\(\left(134-34\right).\left(-28\right)+72.[\left(-55\right)-45]\)

\(=100.\left(-28\right)+72.\left(-100\right)\)

\(=\left(-100\right).\left(28+72\right)\)

\(=-10000\)\(.\)

\(\left(-300\right):20+5.\left(3x-1\right)=25\)

\(\left(-15\right)+5.\left(3x-1\right)=25\)

\(5.\left(3x-1\right)=40\)

\(3x-1=8\)

\(3x=9\)

\(\Rightarrow x=3\)

\(=[\left(-1\right)+\left(-2\right)+3+4]+..+[\left(-97\right)+\left(-98\right)+99+100]\)

\(=4+4+...+4\)    (\(25\) số hạng \(4\))

\(=4.25\)

\(=100\)

\(x+[(-7\)\(03)+12]=-900hay\) \(x+\left(-691\right)=-900\)

Suy ra \(x=\left(-900\right)-\left(-691\right)=-209\)

X+[(-703)+12]=-900
X+(-691)        =-900
X                    =-900-(-691)
X                    =-209
Vậy X=-209