Tính nhanh: E=20203+1/20202-2019
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Cho A = 20203 và B = 2019. 2020. 2021. Không tính cụ thể các giá trị của A và B, hãy so sánh A và B.
2019 nhân 100 thì bằng 201900 > 20203
2020.2021 lớn hơn 100 suy ra b lớn hơn a
\(A=\dfrac{1}{2020}+\dfrac{1}{2020^2}+...+\dfrac{1}{2020^{2021}}\)
\(\Rightarrow2020A=1+\dfrac{1}{2020}+...+\dfrac{1}{2020^{2020}}\)
\(\Rightarrow2020A-A=\left(1+\dfrac{1}{2020}+...+\dfrac{1}{2020^{2020}}\right)-\left(\dfrac{1}{2020}+\dfrac{1}{2020^2}+...+\dfrac{1}{2020^{2021}}\right)\)
\(\Rightarrow2019A=1-\dfrac{1}{2020^{2021}}< 1\Rightarrow A< \dfrac{1}{2019}\)
=10101x5/10101-10101x5/20202+10101x5/30303-10101x5/40404
=5-5/2+5/3-5/4
=35/12
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:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)
ta có :\(E=\frac{2019^{2019}+1}{2019^{2020}+1}\Leftrightarrow2019\cdot E=\frac{2019^{2020}+2019}{2019^{2020}+1}=1+\frac{2019}{2019^{2020}+1}\)
\(F=\frac{2019^{2020}+1}{2019^{2021}+1}\Leftrightarrow2019\cdot F=\frac{2019^{2021}+2019}{2019^{2021}+1}=1+\frac{2019}{2019^{2021}+1}\)
vì \(\frac{2019}{2019^{2020}+1}>\frac{2019}{2019^{2021}+1}\) nên E>F
E=2019 x 2019 x 2019 x ........ x 2019 x2019 +1 /2019 x 2019 x 2019 x.........x 2019 x 2019 + 1
E=1+1/2019+1
E=2/2020
E=1/1010
F=2019 x 2019 x 2019 x .......... x 2019 x 2019 +1 / 2019 x 2019 x 2019 x ....... x 2019 x 2019 +1
F= 1+1/2019+1
F=2/2020
F=1/1010
từ đó ta có E=F(=1/1010)
10101x(5/10101-5/20202+5/30303-5/40404)=10101x5/10101-10101x5/10101x2+10101x5/10101x3-10101x5/10101x4
=5-5/2+5/3-5/4
=35/12
k mk nha
a)( -26) +13+ 10+ 26
=(-26+26)+(13+10)
=0+23
=23
b)17+ 33+ ( -27)+ ( -33)
=[33+(-33)]+[17+(-27)]
=0+(-10)
=-10
c)( -12) +( -250)+ (-8) +250
=[-12+(-8)]+[-250+250]
=-20+0
=-20
d)( -19)+ ( -11) +31 +( -1)
=[-19+(-1)]+[-11+31]
=-20+20
=0
e)(2019- 96)- 2019
=2019-96-2019
=(2019-2019)-96
=0-96
=-96
Bài 1. Tính nhanh các tổng sau
a) (-26) +13 + 10+ 26
= [(-26) + 26] + (13 + 10)
= 0 + 23
= 23
b) 17 + 33 + (-27) + (-33)
= [17 + (-27)] + [33 + (-33)]
= -10 + 0
= -10
c) (-12) + (-250) + (-8) + 250
= [(-12) + (-8)] + [(-250) + 250]
= -20 + 0
= -20
d) (-19) + (-11) + 31 +(-1)
= [(-19) + (-11)] + [31 +(-1)]
= -30 + 30
= 0
e) (2019 - 96) - 2019
= 2019 - 96 - 2019
= (2019 - 2019) - 96
=0 - 96
#học tốt
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