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

a)\(36^2+26^2-52.36\)

=\(\left(36-26\right)^2\)

=\(10^2\)

=100

b)\(99^3+1+3\cdot\left(99^2+99\right)\)

\(=99^3+\left(3\cdot99^2\cdot1\right)+\left(3\cdot99\cdot1^2\right)+1^3\)

\(=\left(99+1\right)^3=100^3=100000\)

a) 39601.              b) 9801.             c) 249001.            d) 89999.

a,  

\(\frac{14}{6}+\frac{1}{9}+\frac{19}{13}+\frac{17}{9}+\frac{7}{13}+\frac{4}{6}\)

\(=\left(\frac{14}{6}+\frac{4}{6}\right)+\left(\frac{1}{9}+\frac{17}{9}\right)+\left(\frac{19}{13}+\frac{7}{13}\right)\)

\(=\frac{18}{6}+\frac{18}{9}+\frac{26}{13}\)

  \(=3+2+2\)

\(=7\)

b,

\(\frac{995}{997}x\frac{990}{993}x\frac{997}{990}x\frac{993}{995}x\frac{97}{95}\)

\(=\frac{995x990x997x993x97}{997x993x990x995x95}\)
\(=\frac{97}{95}\)

(Cùng triệt tiêu 995 ; 990 ; 997 ; 993 )

a)=1+(2-3-4+5)+(6-7-8+9)+...+(994-995-996+997)+998

=1+0+0+...+0+998

=999

b)=(1-3)+(5-7)+...+(97-99)+101

=-2+(-2)+...+(-2)+101                  (50 số hạng -2)

=-2.50+101

=-100+101

=1

a) = (2-3-4+5) + ..... + (993 - 994 - 995 + 996)  + 997 + 998 + 1

= 998 + 997 + 1 = 1996

1+2-3-4+5+6-7-8+...........+993+994-996+997+998

=1+(2-3-4+5)+(6-7-8+9)+...........+(994-996+997+998)

=1+0+0+...........+?

=>Sai đề

b,1-3+5-7+..........+97-99+101

=(-2)+(-2)+..........+(-2)+101(50 số-2)

=(-2).50+101

=-100+101=1