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Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
a) \(52^2\)
\(=\left(50+2\right)^2\)
\(=50^2+2\cdot2\cdot50+2^2\)
\(=2500+200+4\)
\(=2704\)
b) \(98^2\)
\(=\left(100-2\right)^2\)
\(=100^2-2\cdot100\cdot2+2^2\)
\(=10000-400+4\)
\(=9604\)
`a, 52^2 = (50+2)^2 = 2500 + 200 + 4 = 2704`
`b, 98^2 = (100-2)^2 = 100^2 - 2 . 100 . 2 + 4 = 10000 - 400 + 4`
`= 9604`
a) \(82\cdot78\)
\(=\left(80+2\right)\cdot\left(80-2\right)\)
\(=80^2-2^2\)
\(=6400-4\)
\(=6396\)
b) \(87\cdot93\)
\(=\left(90-3\right)\left(90+3\right)\)
\(=90^2-3^2\)
\(=8100-9\)
\(=8091\)
c) \(125^2-25^2\)
\(=\left(125-25\right)\left(125+25\right)\)
\(=100\cdot150\)
\(=15000\)
`a, 38 . 42 = (40-2)(40+2) = 1600 - 4 = 1596`.`
`b, 102^2 = (100+2)^2 = 10000 + 400 + 4 = 10404`
`c, 198^2 = (200-2)^2 = 40000 - 800 + 4 = 39204`
`d, 75^2 - 25^2 = (75+25)(75-25) = 100. 50 = 5000`
3A:
a: \(75\cdot20.9+5^2\cdot20.9\)
\(=20.9\cdot100\)
=2090
b: \(86\cdot15+150\cdot1.4\)
\(=86\cdot15+15\cdot14\)
\(=15\cdot100=1500\)
c: \(93\cdot32+14\cdot16\)
\(=93\cdot32+32\cdot7\)
=3200
\(a.75^2-50.75+25^2\\ =75^2-2.25.75+25^2\\ =\left(75-25\right)^2=50^2=2500\)
\(b.103.97\\ =\left(100+3\right)\left(100-3\right)\\ =100^2-3^2\\ =9991\)
\(a,75^2-50.75+25^2\\ =75^2-2.75.25+25^2\\ =\left(75+25\right)^2=100^2=10000\\ b,103.97\\ =\left(100+3\right).\left(100-3\right)\\ =100^2-3^2=10000-9=9991\)
a. Ta có: \(17^2-14.17+49=17^2-2.7.17+7^2=\left(17-7\right)^2=10^2=100\)
b. \(2021^2-2020^2=\left(2021-2020\right)\left(2021+2020\right)=4041\)
\(101^2=\left(100+1\right)^2=10000+200+1=10201\\ 9999^2=\left(10000-1\right)^2=100000000-20000+1=99980001\\ 47\cdot53=\left(50-3\right)\left(50+3\right)=2500-9=2491\\ 991\cdot1009=\left(1000-9\right)\left(1000+9\right)=1000000-81=999919\)
a: \(101^2=10201\)
b: \(9999^2=99980001\)
c: \(47\cdot53=2491\)
d: \(991\cdot1009=999919\)
a) \(153^2-53^2=\left(153-53\right)\left(153+53\right)=100.206=20600\)
b)
\(\left(2020^2-2019^2\right)+\left(2018^2-2017^2\right)+...+\left(2^2-1^2\right)\\ =\left(2020+2019\right)\left(2020-2019\right)+\left(2018+2017\right)\left(2018-2017\right)+...+\left(2+1\right)\left(2-1\right)\\ =2020+2019+2018+2017+...+2+1\\ =\dfrac{\left(2020+1\right)2020}{2}=2041210\)
Lời giải:
a. $153^2-53^2=(153-53)(153+53)=100.206=20600$
b.
$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$
$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$
$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$
$=2020+2019+2018+2017+...+2+1$
$=\frac{2020.2021}{2}=2041210$
a) = (50-1)^2 = 50^2 - 2.50 + 1 = 2500 - 100 + 1 = 2401
b) = (50+1)^2 = 50^2 + 2.50 + 1 = 2601