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)
So sánh A và B, biết: A= \(\dfrac{2019x2020}{2019x2020+1}\) và B= \(\dfrac{2020}{2019}\)
Ta có :
\(A=\dfrac{2019\times2020}{2019\times2020+1}=\dfrac{2019\times2020+1-1}{2019\times2020+1}=1-\dfrac{1}{2019\times2020+1}\)
Suy ra A < 1 (1)
Lại có \(B=\dfrac{2020}{2019}=\dfrac{2019+1}{2019}=\dfrac{2019}{2019}+\dfrac{1}{2019}=1+\dfrac{1}{2019}\)
Suy ra B > 1 (2)
Từ (1) và (2) ta có : A < 1 < B
=> A < B
Vậy A < B
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
=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)
So sánh các phân số :
A) 2019x2020
2019x2020+1
B) 2020
2021
? x ? = ?
=?
= ? : ?
=...................
hok tốt :)))
(LAUGH) :)))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))))
hok tốt vcl
1/5x6+1/6x7+1/7x8+...+1/2019x2020
1/5x6+1/6x7+1/7x8+...+1/2019x2020
= 6/5+7/6+8/7+...+2020/2019
Rút gọn cho nhau ta còn 2020/5=404
1/5 x 6 + 1/6 x7 + 1/7 x8 + ... + 1/2019 - 1/ 2020
=1/5 -1/6 +1/6 -1/7 + 1/7 - 1/8 + ... + 1/2019 - 1/2020
Sau khi giản ước, ta còn:
1/5 - 1/2020 = 403/2020.
Đáp số: 403/2020
Mình xin lôĩ, mình nhìn nhầm đề mong bạn thông cảm.
tính nhanh 2018x2019+4036/2019x2020-2
Ta có \(\frac{2018\times2019+4036}{2019\times2020-2}\)
\(=\frac{\left(2020-2\right)\times2019}{2019\times2020-2}\)
\(=\frac{2020\times2019-2\times2019+4036}{2019\times2020-2}\)
\(=\frac{2020\times2019-4038+4036}{2019\times2020-2}\)
\(=\frac{2020\times2019-2}{2019\times2020-2}\)
\(=1\)
1/6.7+1/7x8+1/8x9+...+1/2019x2020 Giúp mình pls
Dặt \(A=\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2019.2020}\)
\(A=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=\frac{1}{6}-\frac{1}{2020}\)
\(A=\frac{1007}{6060}\)
hok tốt!!
Tính tổng: 1x2 + 2x3 + 3x4 +...+ 2019x2020 + 2020x2021
Đặt A = 1.2 + 2.3 + 3.4 + ... + 2019.2020 + 2020.2021
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 2019.2020.3 + 2020.2021.3
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2019.2020.(2021 - 2018) + 2020.2021.(2022 - 2019)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2019.2020.2021 - 2018.2019.2020 + 2020.2021.2022 - 2019.2020.2021
=> 3A = 2020.2021.2022
=> A = 2 751 551 080
Đặt \(A=1.2+2.3+3.4+.........+2019.2020+2020.2021\)
\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+.....+2019.2020.3+2020.2021.3\)
\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+.....+2020.2021.\left(2022-2019\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2020.2021.2022-2019.2020.2021\)
\(=2020.2021.2022\)
\(\Rightarrow A=\frac{2020.2021.2022}{3}\)
Có mấy số 0 tận cùng: 2000x2001x2002x2003x...2019x2020
