a)Ta có \(101^2\)=\(\left(100+1\right)^2\)=10000+200+1
=10201
b)\(199^2\)=\(\left(200-1\right)^2=40000-400+1\)=39601
c)47.53=\(\left(50-3\right)\left(50+3\right)=50^2-3^2\)=2500-9=2491
giải
(47.53)=(47+3)(53-3)
=50.50=2500
a) 1012=(100+1)=1002+2.100.1+12=10000+200+1=10201
b)1992=(200-1)=2002-2.200.1+12=40000-400+1=39601
c)47.53=(50-3)(50+3)=502-32=2500-9=2491
a) \(101^2=\left(100+1\right)^2=100^2+2.100.1+1^2=10201\)
b) \(199^2=\left(100+99\right)^2=100^2+2.100.99+99^2=39601\)
c) \(47.53=\left(50-3\right)\left(50+3\right)=50^2-3^2=2491\)
a) 1012=(100+1)2=10000+200+1=10201
b) 1992=(200-1)2=40000-400+1=39601
c)47.53=(50-3)(50+3)=2500-9=2491