Tính nhanh:
a,99^2
b, 2001.1999
c,48^2 +4.48 +4
d,101^2 - 202+1
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\(101^2=\left(100+1\right)^2=10000+200+1=10201\\ 9999^2=\left(10000-1\right)^2=100000000-20000+1=99980001\\ 47\cdot53=\left(50-3\right)\left(50+3\right)=2500-9=2491\\ 991\cdot1009=\left(1000-9\right)\left(1000+9\right)=1000000-81=999919\)
a: \(101^2=10201\)
b: \(9999^2=99980001\)
c: \(47\cdot53=2491\)
d: \(991\cdot1009=999919\)
a: \(19\cdot35+55\cdot81+19\cdot20\)
\(=19\cdot55+55\cdot81\)
\(=55\cdot100=5500\)
b: \(117\cdot98-117\cdot46-52\cdot17\)
\(=117\cdot52-52\cdot17\)
=5200
c: Số số hạng là:
\(\left(50-2\right):2+1=48:2+1=25\left(số\right)\)
Tổng là:
\(\dfrac{\left(50+2\right)\cdot25}{2}=\dfrac{52\cdot25}{2}=650\)
d: Số số hạng là:
50-11+1=40(số)
Tổng là:
\(\dfrac{\left(50+11\right)\cdot40}{2}=61\cdot20=1220\)
a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)
\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)
\(=\dfrac{2}{5}+\dfrac{4}{2}\)
\(=\dfrac{2}{5}+2\)
\(=\dfrac{2}{5}+\dfrac{10}{5}\)
\(=\dfrac{12}{5}\)
b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)
\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)
\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)
\(=1-\dfrac{2}{2008}\)
\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)
\(=\dfrac{2006}{2008}\)
\(=\dfrac{1003}{1004}\)
a: =2/5+4/2
=2/5+2
=12/5
b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)
\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)
a) \(99^2=\left(100-1\right)^2=100^2-2.100+1=9801\)
b) \(2001.1999=\left(2000+1\right)\left(2000-1\right)=2000^2-1^2=3999999\)
c) \(48^2+4.48+4=\left(48+2\right)^2=50^2=2500\)
d) \(101^2-202+1=101^2-2.101+1=\left(101-1\right)^2=100^2=10000\)