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

bằng 1

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

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

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

\(=1\)

Ta có |x+3| >=0 với mọi x

=> |x-3|+11 >=11 hay A >=11

Dấu "=" xảy ra <=> |x-3|=0

<=> x-3=0

<=> x=3

Vậy Min_A=11 đạt được khi x=3

a, 4*113*25-5*112*20=(4*25)*113-(5*20)*112=100*113-100*112=100*(113-112)=100

b, 2+4+...........+98+100-(1+3+...........+99)=[(100-2)/2+1*102]-[(99-1)/2+1*100]

 =98/2+1*102-98/2+1*100=(98/2+1)*(102-100)=(49+1)*2=50*2=100

4x113x25-5x112x20

= ( 4x25 )x113-(5x20)x112

=100x113-100x112

=100x(113-112)

=100x1

=100

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{63}+\frac{100}{99}\\ =\frac{2.2}{1.3}+\frac{4.4}{3.5}+\frac{6.6}{5.7}+\frac{8.8}{7.9}+\frac{10.10}{9.11}\)

\(\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{65}+\frac{100}{99}\)

\(1+\frac{1}{3}+1+\frac{1}{15}+1+\frac{1}{35}+1+\frac{1}{65}+1+\frac{1}{99}\)

\(\left(1+1+1+1+1\right)+\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{65}+\frac{1}{99}\right)\)

\(\frac{60}{11}\)

\(\frac{3}{25}-\frac{11}{4}+\frac{-28}{25}-\frac{4}{3}-\frac{1}{4}\)

\(=\left(\frac{3}{25}+\frac{-28}{25}\right)+\left(\frac{11}{4}-\frac{1}{4}\right)-\frac{4}{3}\)

\(=\frac{25}{25}+\frac{5}{2}-\frac{4}{3}\)

\(=1+\frac{5}{2}-\frac{4}{3}\)

\(=\frac{13}{6}\)