(2021×2019-185)/( 3857+2021×2017) giúp mình nhé
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Xét khai triển:
\(2^{2021}=\left(1+1\right)^{2021}=C_{2021}^0+C_{2021}^1+...+C_{2021}^{2020}+C_{2021}^{2021}\) (1)
\(0=\left(1-1\right)^{2021}=C_{2021}^0-C_{2021}^1+C_{2021}^2+...+C_{2021}^{2020}-C_{2021}^{2021}\) (2)
Trừ vế cho vế (1) và (2):
\(2^{2021}=2.C_{2021}^1+2.C_{2021}^3+...+2C_{2021}^{2021}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+C_{2021}^{2021}=\dfrac{2^{2021}}{2}=2^{2020}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}+1=2^{2020}\)
\(\Rightarrow C_{2021}^1+C_{2021}^3+...+C_{2021}^{2019}=2^{2020}-1\)
2021 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
2021 x 2021 - 2019 x 2023
= (2019 +2) x ( 2023 -2) - 2019 x 2023
= 2019 x 2023 - 2 x 2019 + 2 x 2023 - 4 - 2019 x 2023
= ( 2019 x 2023 - 2019 x 2023) + 2 x ( 2023 - 2019) - 4
= 0 + 2 x 4 - 4
= 8 - 4
= 4
2019 . x + 1/2021 . x + 1/2023 . x - 1/2023 = 2019 + 1/2021
mọi người ơi trả lời nhanh giùm mình nhé
\(\dfrac{2021}{2019}=1+\dfrac{2}{2019}\)
2019/2017=1+2/2017
mà 2/2019<2/2017
nên 2021/2019<2019/2017
\(\dfrac{x-2017}{2019}+\dfrac{x-2019}{2017}=\dfrac{x+6}{2021}\)
\(\Rightarrow\dfrac{x-2017}{2019}-1+\dfrac{x-2019}{2017}-1=\dfrac{x+6}{2021}-2\)
\(\Rightarrow\dfrac{x-2017}{2019}-\dfrac{2019}{2019}+\dfrac{x-2019}{2017}-\dfrac{2017}{2017}=\dfrac{x+6}{2021}-\dfrac{4042}{2021}\)
\(\Rightarrow\dfrac{x-2017-2019}{2019}+\dfrac{x-2019-2017}{2017}=\dfrac{x+6-4042}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2019}+\dfrac{x-4036}{2017}=\dfrac{x-4036}{2021}\)
\(\Rightarrow\dfrac{x-4036}{2021}-\dfrac{x-4036}{2019}-\dfrac{x-4036}{2017}=0\)
\(\Rightarrow\left(x-4036\right)\left(\dfrac{1}{2021}-\dfrac{1}{2019}-\dfrac{1}{2017}\right)=0\)
=> x - 4036 = 0
=> x = 4036
x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 + x−2019/2017 = x+6/2021
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 1 + x − 2019/2017 − 1 = x + 6/2021 − 2
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017/2019 − 2019/2019 + x − 2019/2017 − 2017/2017
= x + 6/2021 − 4042/2021
=> x − 2017 − 2019/ 2019 + x − 2019 − 2017/2017
= x + 6 − 4042/2021
=> x − 2017 − 2019/2019 + x − 2019 − 2017/2017 = x + 6 − 4042/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2019 + x − 4036/2017 = x − 4036/2021
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=> x − 4036/2021 − x − 4036/2019 − x − 4036/2017 = 0
=>(x − 4036)(12021 − 12019 − 12017) = 0
=> x - 4036 = 0
=> x = 4036
A = 2021/2022+2020/2021+2019/2020+2018/2019+2017/2018
A<2022/2022+2021/2021+2020/2020+2019/2019+2018/2018
A<1+1+1+1+1
A<5
2017/2020<2019/2020< 1
1< 2022/2021< 2023/2021
vậy phân số lớn nhất là 2023/2021
ta so sánh với 1:
2017/2020<2019/2020< 1
1< 2022/2021< 2023/2021
nên phân số lớn nhất là phân số cuối: 2023/2021
Ta có:\(A=\frac{2021x2019-185}{3857+2021x2017}=\frac{2021x2019-185}{4042+2021x2017-185}=\frac{2021x2019-185}{2021x2019-185}=1\)
Đây là tính nhanh
hay
tính bình thường thế bạn ?