giúp e 2 bài này với ạ
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ND
2
10 tháng 7 2023
5:
a: =>x^2-2x+1<x^2+3x
=>-5x<-1
=>x>1/5
b: =>x^2-4>x^2-4x
=>-4x<-4
=>x>1
c: =>2x+3=6-3+4x=4x+3
=>-2x<0
=>x>0
d: =>-7x-2>2x+3-5+6x=8x-2
=>-15x>0
=>x<0
NL
0
DT
Đỗ Thanh Hải
CTVVIP
29 tháng 6 2021
A
1 were having
2 wasn't listening
3 was lying - was eating
4 were smiling - was taking
5 were - arguing
6 was talking
7 was - chatting
8 wasn't sleeping
B
1 was raining
2 was wearing
3 carrying
4 were leaving
5 were hurrying
6 wasn't wearing
7 was standing
8 was - following
mn giúp e 2 bài này với ạ
H
5 tháng 1 2022
bạn đăng tách ra tầm 10 câu mỗi lần đăng nha, chứ dài ntnay ngại làm lắm~
5 tháng 1 2022
e có tách 3 bài ra rồi ạ, phiền anh/chị/bạn giúp e với ạ, e cảm ơn ạ
LT
0
@ Nguyễn Tuấn Tú bạn không nên bỏ bước phá ngoặc nhé!
Bài 2:
\(H=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{2017}{3^{2017}}\\ \Rightarrow3H=1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{2017}{3^{2016}}\\ \Rightarrow3H-H=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+...+\dfrac{2017}{3^{2016}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+\dfrac{4}{3^4}+...+\dfrac{2017}{3^{2017}}\right)\\ \Rightarrow2H=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}-\dfrac{2017}{3^{2017}}\)
Đặt \(A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}\)
\(⇒3A=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2015}} \)
\(⇒3A-A=(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{2015}})-(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2016}}) \)
\(\Rightarrow2A=3-\dfrac{1}{3^{2016}}\\\Rightarrow A=\dfrac{3-\dfrac{1}{3^{2016}}}{2} \)
Thay vào 2H được:
\(2H=\dfrac{3-\dfrac{1}{3^{2016}}}{2}-\dfrac{2017}{3^{2017}}\\ \Rightarrow H=\dfrac{\dfrac{3-\dfrac{1}{3^{2016}}}{2}-\dfrac{2017}{3^{2017}}}{2}=\dfrac{3-\dfrac{1}{3^{2016}}}{4}-\dfrac{2017}{2.3^{2017}}\)
Vậy...