cho A=2+2^2+2^3+.......+2^2024+2^2025❤
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
LN
0
AT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9 2023
A = \(\dfrac{1}{1+2+3}\)+\(\dfrac{1}{1+2+3+4}\)+...+ \(\dfrac{1}{1+2+...+2004}\)+ \(\dfrac{2}{2025}\)
A = \(\dfrac{1}{\left(1+3\right).3:2}\)+\(\dfrac{1}{\left(4+1\right).4:2}\)+...+ \(\dfrac{1}{\left(2024+1\right).2024:2}\)+\(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3.4}\)+\(\dfrac{2}{4.5}\)+...+\(\dfrac{2}{2024.2025}\)+ \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+...+ \(\dfrac{1}{2024.2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)+...+ \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = 2.(\(\dfrac{1}{3}\) - \(\dfrac{1}{2025}\)) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\) - \(\dfrac{2}{2025}\) + \(\dfrac{2}{2025}\)
A = \(\dfrac{2}{3}\)
5 tháng 6 2017
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}].[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}]}\)
=\(\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)
=\(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Áp dụng ta có S=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
15 tháng 10 2018
Ta có công thức tổng quát:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{2025\sqrt{2024}+2024\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{1}{\sqrt{4}}+...+\dfrac{1}{\sqrt{2024}}-\dfrac{1}{\sqrt{2025}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2025}}=1-\dfrac{1}{45}=\dfrac{44}{45}\)
NA
9 tháng 1
a, 2\(^3\) . x + 2005\(^0\) . x = 994-15:3+1\(^{2025}\)
8 .x + 1 . x = 990
x . [ 8 +1 ] = 990
x . 9 = 990
x = 990 : 9
x = 110
10 tháng 1
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5
DN
18 tháng 5 2023
Đề có phải là:
\(\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}=4\text{ ?}\)
\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-4=0\)
\(\Rightarrow\text{ }\dfrac{x+1}{2024}+\dfrac{x+2}{2025}+\dfrac{x+3}{2026}+\dfrac{x+4}{2027}-1-1-1-1=0\)
\(\Rightarrow\left(\dfrac{x+1}{2024}-1\right)+\left(\dfrac{x+2}{2025}-1\right)+\left(\dfrac{x+3}{2026}-1\right)+\left(\dfrac{x+4}{2027}-1\right)=0\)
\(\Rightarrow\left(\dfrac{x+1-2024}{2024}\right)+\left(\dfrac{x+2-2025}{2025}\right)+\left(\dfrac{x+3-2026}{2026}\right)+\left(\dfrac{x+4-2027}{2027}\right)=0\)
\(\Rightarrow\dfrac{x-2023}{2024}+\dfrac{x-2023}{2025}+\dfrac{x-2023}{2026}+\dfrac{x-2023}{2027}=0\)
\(\Rightarrow\left(x-2023\right)\left(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\right)=0\)
Mà \(\dfrac{1}{2024}+\dfrac{1}{2025}+\dfrac{1}{2026}+\dfrac{1}{2027}\ne0\)
\(\Rightarrow x-2023=0\)
\(\Rightarrow x=0+2023\)
\(\Rightarrow x=2023\)
Vậy, \(x=2023.\)
HL
1
8 tháng 2 2023
\(1:\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{2024}{2025}\)
= \(1\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2025}{2024}=\dfrac{2025}{2}\)
HV
3
NP
3
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 11 2023
c, |2\(x\) + 1| + |3\(x\) - 1| = 0
vì |2\(x\) + 1| ≥ 0; |3\(x\) - 1| = 0
⇒ |2\(x\) + 1| + |3\(x\) - 1| = 0
⇔ \(\left\{{}\begin{matrix}2x+1=0\\3x-1=0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}2x=-1\\3x=1\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(-\dfrac{1}{2}\) < \(\dfrac{1}{3}\)
Vậy \(x\) \(\in\) \(\varnothing\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
22 tháng 11 2023
a, Nếu 4.|3\(x\) - 1| = |6\(x\) - 2| + |-1,5|
4.|3\(x\) -1| - 2.|3\(x\) - 1| = 1,5
Nếu 3\(x\) - 1 ≥ 0 ⇒ \(x\) ≥ \(\dfrac{1}{3}\)
Ta có: 4.(3\(x\) - 1) - 2.(3\(x\) - 1) = 1,5
12\(x\) - 4 - 6\(x\) + 2 = 1,5
6\(x\) - 2 = 1,5
6\(x\) = 1,5 + 2
6\(x\) = 3,5
\(x\) = 3,5: 6
\(x\) = \(\dfrac{7}{12}\)
Nếu 3\(x\) - 1 < 0 ⇒ \(x\) < \(\dfrac{1}{3}\)
Ta có: - 4.(3\(x\) - 1) = - (6\(x\) - 2) + 1,5
-12\(x\) + 4 + 6\(x\) - 2 = 1,5
-6\(x\) + 2 = 1,5
6\(x\) = 2- 1,5
6\(x\) = 0,5
\(x\) = 0,5 : 6
\(x\) = \(\dfrac{1}{12}\)
Vậy \(x\) \(\in\) {\(\dfrac{1}{12}\); \(\dfrac{7}{12}\)}
đề bài là gì vậy bn
\(A=2+2^2+2^3+....+2^{2024}+2^{2025}\)
\(\Leftrightarrow2A=2^2+2^3+2^4+....+2^{2025}+2^{2026}\)
\(\Leftrightarrow2A-A=A=2^{2026}-2\)