191919+20202 ____________ 353535+70707
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x0x0x : X x 3 – 20202
= 10101 x 3 – 20202
= 30303 – 20202
= 10101
a. Ta có: \(17^2-14.17+49=17^2-2.7.17+7^2=\left(17-7\right)^2=10^2=100\)
b. \(2021^2-2020^2=\left(2021-2020\right)\left(2021+2020\right)=4041\)
Đặt \(A=\frac{3\cdot7\cdot13\cdot37\cdot39-10101}{505050-70707}\)
\(A=\frac{\left(3\cdot7\cdot13\cdot37\right)\cdot39-10101\cdot1}{50\cdot10101+7\cdot10101}\)
\(A=\frac{10101\cdot39-10101\cdot1}{10101\cdot\left(50+7\right)}\)
\(A=\frac{10101\cdot\left(39-1\right)}{10101\cdot57}\)
\(A=\frac{10101\cdot38}{10101\cdot57}\)
\(A=\frac{38}{57}=\frac{38:19}{57:19}=\frac{2}{3}\)
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) 544544 – 444444 = 544.1001 – 444.1001 = 1001.(544 – 444) = 1001.100 = 100100
b) 131313 – 10101 – 20202 = 10101.13 – 10101 – 10101.2
= 10101.(13 – 1 – 2) = 10101.10 = 101010
so sánh hả bạn?
ta có : 191919 + 20202 = 212121
353535 + 70707 = 424242
vì 212121 < 424242
=> 191919 + 20202 < 353535 + 70707