tìm x biết:1+(x-1)^2+(x-1)^4+...+(x-1)^2020=17^2022-1/((x-1)^2-1)
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
NH
5 tháng 10 2023
\(1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\left(đk:x>2\right)\)
đặt
\(A=1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\)
\(\left(x-1\right)^2A=\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\)
\(\left(x-1\right)^2A-A=\left[\left(x-1\right)^2+\left(x-1\right)^4+\left(x-1\right)^6+...+\left(x-1\right)^{2022}\right]-\left[1+\left(x-1\right)^2+\left(x-1\right)^4+...+\left(x-1\right)^{2020}\right]\)
\(\left[\left(x-1\right)^2-1\right]A=\left(x-1\right)^{2022}-1\)
\(A=\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}\)
\(=>\dfrac{\left(x-1\right)^{2022}-1}{\left(x-1\right)^2-1}=\dfrac{17^{2022}-1}{\left(x-1\right)^2-1}\\ =>\left(x-1\right)^{2022}-1=17^{2022}-1\\ =>\left(x-1\right)^{2022}=17^{2022}\\ =>x-1=17\\ =>x=18\left(tm\right)\)
HK
16 tháng 4 2023
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
MN
1
11 tháng 2 2022
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{505}{1011}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1010}{1011}\)
=>1/x+1=-1009/2022
=>x+1=-2022/1009
hay x=-3031/1009
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
13 tháng 12 2023
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
LL
2
21 tháng 7 2023
(x-1)2020=(x-1)2022
=>(x-1)2020-(x-1)2022=0
=>(x-1)2020-(x-1)2020.(x-1)2=0
=>(x-1)2020(1-(x-1)2=0
=>(x-1)2020=0 hoặc 1-(x-1)2=0
=>x=1 hoặc x=2.
Bài 2
a,2105 và 545
2105=(27)15=12815
545=(53)15=12515
Vì 12815>12515 nên 2105>545.
b,
554 và 381
554=(56)9=156259
381=(39)9=196839
Vì 156259<196839 nên 554<381
ND
Nguyễn Đức Trí
VIP
21 tháng 7 2023
Bài 1 :
\(\left(x-1\right)^{2020}=\left(x-1\right)^{2022}\)
\(\Rightarrow\left(x-1\right)^{2022}-\left(x-1\right)^{2020}=0\)
\(\Rightarrow\left(x-1\right)^{2020}\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=1\\x-1=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
I
5 tháng 1 2023
\(\dfrac{x+1}{2023}+\dfrac{x+2}{2022}=\dfrac{x+3}{2021}+\dfrac{x+4}{2020}\\ \Leftrightarrow\dfrac{x+1}{2023}+1+\dfrac{x+2}{2022}+1=\dfrac{x+3}{2021}+1+\dfrac{x+4}{2020}+1\\ \Leftrightarrow\dfrac{x+1+2023}{2023}+\dfrac{x+2+2022}{2022}-\dfrac{x+3+2021}{2021}-\dfrac{x+4+2020}{2020}=0\\ \Leftrightarrow\left(x+2024\right)\times\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)=0\\ \Rightarrow x+2024=0:\left(\dfrac{1}{2023}+\dfrac{1}{2022}-\dfrac{1}{2021}-\dfrac{1}{2020}\right)\\ \Rightarrow x+2024=0\\ \Rightarrow x=-2024\)
HM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 9 2023
Câu này cô làm rồi em nhá, em xem phần câu hỏi của tôi ý
BC
5
NH
16 tháng 7 2019
\(\frac{x+4}{2019}+\frac{x+3}{2020}=\frac{x+2}{2021}+\frac{x+1}{2020}\)
\(\Leftrightarrow(\frac{x+4}{2019}+1)+(\frac{x+3}{2020}+1)=(\frac{x+2}{2021}+1)+(\frac{x+1}{2022}+1)\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}=\frac{x+2023}{2021}+\frac{x+2023}{2022}\)
\(\Leftrightarrow\frac{x+2023}{2019}+\frac{x+2023}{2020}-\frac{x+2023}{2021}-\frac{x+2023}{2022}=0\)
\(\Leftrightarrow\left(x+2023\right)\left(\frac{1}{2019}+\frac{1}{2020}-\frac{1}{2021}-\frac{1}{2020}\right)=0\)
\(\Leftrightarrow x+2023=0\)
\(\Leftrightarrow x=-2023\)
HM
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
24 tháng 9 2023
Q = \(\dfrac{1+x^4+x^8+...+x^{2020}}{1+x^2+...+x^{2022}}\)
Đặt A = 1 + \(x^4\) + \(x^8\) +...+ \(x^{2020}\)
Đặt B = 1 + \(x^2\) + ...+ \(x^{2022}\)
Thì Q = \(\dfrac{A}{B}\)
A = 1 + \(x^4\) + \(x^8\) + ...+ \(x^{2020}\)
A.\(x^4\) = \(x^4\) + \(x^8\) +....+ \(x^{2020}\) + \(x^{2024}\)
A.\(x^4\) - A = \(x^{2024}\) - 1
A = \(\dfrac{x^{2024}-1}{x^4-1}\)
B = 1 + \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\)
B.\(x^2\) = \(x^2\) + \(x^4\) +...+ \(x^{2020}\) + \(x^{2022}\) + \(x^{2024}\)
B\(x^2\) - B = \(x^{2024}\) - 1
B = \(\dfrac{x^{2024}-1}{x^2-1}\)
Q = \(\dfrac{\dfrac{x^{2024}-1}{x^4-1}}{\dfrac{x^{2024}-1}{x^2-1}}\)
Q = \(\dfrac{x^{2024}-1}{x^4-1}\) \(\times\)\(\dfrac{x^2-1}{x^{2024}-1}\)
Q = \(\dfrac{1}{x^2+1}\)
ai biết làm ko