a, \(\dfrac{2017.2021-4031}{2020+2017.2018}\)
= \(\dfrac{2017\left(2018+3\right)-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2017.3-4031}{2020+2017.2018}\)
= \(\dfrac{2017.2018+2020}{2020+2017.2018}\)
= 1
@Nguyen Thi Ngoc Linh
BT1: So sánh:
1) \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}\) VỚI 3
BT1: Tìm x, biết:
2) \(\dfrac{x+2017}{x+2018}=\dfrac{2020}{2021}\)
so sánh:
A=\(\dfrac{2017^{2017}+1}{2017^{2018}+1}\)và B=\(\dfrac{2017^{2018}-2}{2017^{2019}-2}\)
Tinh nhanh:
2017 2017 2017 x 2018 2018 2018 2018 /2018 2018 2018 x 2017 2017 2017 2017
Tính A+B
A= 1.99+2.98+3.97+.............+99.1
B=1.101+2.102+3.103+............+99.199
so sánh P và Q
P=2016/2017+2017/2018
Q= 2016+2017+2018/2017+2018+2019
BT1: So sánh:
2) \(\dfrac{2017}{2018}+\dfrac{2018}{2019}\) VỚI \(\dfrac{2017+2018}{2018+2019}\)
Tìm x:
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+......+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\)
So sánh A và B
A =\(\frac{2^{2017}+1}{2^{2018}+1}\) và B = \(\frac{2^{2018}+1}{2^{2019}+1}\)
s=1+3+5+...+2017
s=2+4+6+...+2018
s=101+103+105+...+2019
