So sánh A và B, biết: A= \(\dfrac{2019x2020}{2019x2020+1}\) và B= \(\dfrac{2020}{2019}\)
Những câu hỏi liên quan
so sánh 43/52 và 60/120 , So sánh 17/ 68 và 35 / 103 , So sánh 2018 x 2019-1/2018x2019 va 2019x2020-1/2019x2020
a: 43/52>26/52=1/2=60/120
b: 17/68=1/4<1/3=35/105<35/103
c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)
\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)
2018*2019<2019*2020
=>-1/2018*2019<-1/2019*2020
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
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So sánh các phân số :
A) 2019x2020
2019x2020+1
B) 2020
2021
? x ? = ?
=?
= ? : ?
=...................
hok tốt :)))
(LAUGH) :)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
hok tốt vcl
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A=\(\frac{2019x2020}{2019x2020+1}\)và B=\(\frac{2020}{2021}\)
Dấu ''\(x\)'' là dấu nhân chăng ?
\(A=\frac{2019x2020}{2019x2020+1}\)và \(B=\frac{2020}{2021}\)
Bài ra ta có :
Xét \(A=\frac{2019x2020}{2019x\left(2020+1\right)}=\frac{2020}{2020+1}=\frac{2020}{2021}\)
Vì \(\frac{2020}{2021}=\frac{2020}{2021}\)
Suy ra A = B theo (ĐPCM)
Bài 1:
A)
\(\frac{2021x2020-2019x2020}{2019x2020+2021x7+2013}\)
B)
\(\frac{2019x2020+2021x21+1999}{2021x2020-2019x2020}\)
C)
\(\frac{2012x2020+2021+2019}{2021x2020-2019x2020}\)
D)
\(\frac{2021x14+2006+2019x2020}{2020+2020x503+504x2020}\)
AI ĐÓ GIÚP EM VỚI Ạ EM ĐANG CẦN RẤT RẤT LÀ GẤP AAẠA
a=(2021-2019) x 2020/2019x2020+(2020 +1)x7+2013
=1x2020/2019x2020+2020x7+1x7+2013
=2020/(2019+7)x2020+2020
=2020/(2019+1+70) x2020
=2020/2027 x2020
=2020/4112783
Mình cảm ơn ạ nếu bạn có thời gian làm giúp mình câu b c d đc k ạ?:3
so sánh A và B biết:
A=\(\dfrac{2^{2018}}{2^{2018}+3^{2019}}\)+\(\dfrac{3^{2019}}{3^{2019}+5^{2020}}\)+\(\dfrac{5^{2020}}{5^{2020}+2^{2018}}\)
B=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{5.6}\)+...+\(\dfrac{1}{2019.2020}\).
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
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12 tháng 5 2021 lúc 20:40
So sánh : \(A=\dfrac{2019^{2020}+1}{2019^{2019}-1}\) và \(B=\dfrac{2019^{2019}+1}{2019^{2018}-1}\)
Lời giải:
Ta có:
\(A+1=\frac{2019^{2019}+2019^{2020}}{2019^{2019}-1}=\frac{2019^{2019}.2020}{2019^{2019}-1}\)
\(B+1=\frac{2019^{2019}+2019^{2018}}{2019^{2018}-1}=\frac{2019^{2018}.2020}{2019^{2018}-1}\) \(=\frac{2019^{2019}.2020}{2019^{2019}-2019}>\frac{2019^{2019}.2020}{2019^{2019}-1}\)
$\Rightarrow B+1>A+1$
$\Rightarrow B>A$
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1. So sánh
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\) và B= \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{13}{60}\)
b) \(C=\dfrac{2019}{2021}+\dfrac{2021}{2022}\) và \(D=\dfrac{2020+2022}{2019+2021}.\dfrac{3}{2}\)
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
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Trả lời giúp mình 3 câu này nha! AI làm đúng mình sẽ tick cho
a, So sánh các phân số sau: A= 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2019x2020 với 1
b, So sánh các phân số sau : A= 2018/2019 + 2019/2020 + 2020/2018 với 3
c, So sánh các phân số sau F= 1+1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45 với 1/2
Mong các bạn trả lời câu hỏi của mình =>
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)
\(=1-\frac{1}{2020}>1\)
Thank you bạn dcv new ^ ^
nhầm dấu rồi bé hơn 1 chứ:v bạn sửa lại hộ mình nhà
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4 tháng 5 2021 lúc 20:36
So sánh A và B
A=\(\dfrac{4-7^{2020}}{7^{2020}}\)+\(\dfrac{5+7^{2021}}{7^{2021}}\)
B=\(\dfrac{1}{7^{2019}}\)
Ta có:
\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)
\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)
Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)
=> B>A
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