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: \(36^2+26^2-52\cdot36=\left(36-26\right)^2=10^2=100\)

b: \(99^3+1+3\left(99^2+99\right)\)

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

=100^3=10^6

a) 699 991 ; 699 992 ; 699 993 ; 699 994 ; 699 995 ; 699 996

b) 700 007 ; 700 008 ; 700 009 ; 700 010 ; 700 011 ; 700 012

\(A=1-2-3+4+5-6-7+8...+993-994-995+996+997\)

\(A=\left(1-2-3+4\right)+\left(5-6-7+8\right)...+\left(993-994-995+996\right)+997\)

\(A=0+0+...+0+997=997\)

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 )

5830+(-993)+170+(-4007)=1000

5830 + (-993) + 170 + (-4007)

= (5830 + 170) + (-993 - 4007)

= 6000 + (-5000)

= 1000

5830+(-993)+170+(-4007)=1000

\(\left(1981\times1982-990\right):\left(1980\times1982+992\right)\)

\(=\left(1980\times1982+1982-990\right):\left(1980\times1982+992\right)\)

\(=\left(1980\times1982+992\right):\left(1980\times1982+992\right)\) 

\(=1\)