Lời giải:

a. $99^3+1+3(99^2+99)=99^3+3.99^2.1+3.99.1^2+1^3=(99+1)^3=100^3=1000000$

b. $11^3-1-3(11^2-11)=11^3-3.11^2.1+3.11.1^2-1^3=(11-1)^3=10^3=1000$

(102 + 112 + 122) : (132 + 142)

= (100 + 121 + 144) :( 169 + 196)

= 365: 365

= 1 

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n}<1\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1+1=2\)

=>A<2

Ta có: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>0\)

=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}>1\)

=>A>1

Do đó: 1<A<2

=>A không là số tự nhiên

(-245) + (-32) + 45 + 132 + (-8)

=(45 - 245) + (132 - 32) - 8

= -200 + 100 - 8

= -108

112 + (-61) + (-95) + (-12) + (-39)

=(112 - 12) - (61 + 39) - 95

= 100 - 100 - 95

= -95

Áo Lai Giai chó nhai dell rách

132² - 112²

= (132 - 112).(132 + 112)

= 20.244

= 10.2.244

= 10.488

= 4880

400

\(=\left(152-148\right)+\left(144-140\right)+\left(136-132\right)+\left(128-124\right)+\left(120-116\right)+\left(112-108\right)+\left(104-100\right)\\ =4+4+4+4+4+4+4=28\)

=(152−148)+(144−140)+(136−132)+(128−124)+(120−116)+(112−108)+(104−100)=4+4+4+4+4+4+4=28