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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.
$2020^2-2019^2+2018^2-2017^2+...+2^2-1^2$
$=(2020^2-2019^2)+(2018^2-2017^2)+...+(2^2-1^2)$
$=(2020-2019)(2020+2019)+(2018-2017)(2018+2017)+...+(2-1)(2+1)$
$=2020+2019+2018+2017+...+2+1$
$=\frac{2020.2021}{2}=2041210$
Ta có A = 2019.2021.a = (2020 – 1)(2020 + 1)a = ( 2020 2 – 1)a
Và B = ( 2019 2 + 2 . 2019 + 1 ) a = ( 2019 + 1 ) 2 a = 2020 2 a
Vì 2020 2 – 1 < 2020 2 và a > 0 nên ( 2020 2 – 1 ) a < 2020 2 a hay A < B
Đáp án cần chọn là: D
Ta có A = 2016.2018.a = (2017 – 1)(2017 + 1)a = ( 2017 2 – 1)a
Vì 2017 2 – 1 < 2017 2 và a > 0 nên 2017 2 – 1 a < 2017 2 a hay A < B
Đáp án cần chọn la: B
SSH:(20152-12):10+1=2015
(12-22)+(32-42)+(52-62)+...+(20132-20142)+20152
-10+(-10)+(-10)+...+(-10)+20152
-10x(2015-1):2+20152=12
=> C=12
có: \(2020.2022=\left(2021-1\right)\left(2021+1\right)=2021^2-1< 2021^2\)
a. Ta có: \(17^2-14.17+49=17^2-2.7.17+7^2=\left(17-7\right)^2=10^2=100\)
b. \(2021^2-2020^2=\left(2021-2020\right)\left(2021+2020\right)=4041\)
1.
$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$
2.
$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$
3. Không phù hợp để tính nhanh
4.
$=15^8-(15^8-1)=1$
5.
$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$
$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$
$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$
6:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)