Tìm n
a, 4 ⁿ+² = 64
b, (x+2)⁴ = 625
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
PT
2
L
26 tháng 12 2019
(x-1)3\(=\)(x-5)3
\(\Leftrightarrow\)(x-1)3-(x-1)5\(=\)0
\(\Leftrightarrow\)(x-1)3\([\)1-(x-1)2\(]\)\(=\)0
\(\Leftrightarrow\)(x-1)3\(=\)0 hoặc 1-(x-1)2\(=\)0
\(\Leftrightarrow\)x-1\(=\)0 hoặc x-1\(=\pm\)1
\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0
Vậy x\(\in\){1;2;0}
b) (x-1)n\(=\)(x-1)n+2
\(\Leftrightarrow\)(x-1)n-(x-1)n+2\(=\)0
\(\Leftrightarrow\)(x-1)n\([\)1-(x-1)2\(]\)\(=\)0
\(\Leftrightarrow\)(x-1)n\(=\)0 hoặc (x-1)2\(=\)1
\(\Leftrightarrow\)x\(=\)1 hoặc x\(=\)2; x\(=\)0
Vậy x\(\in\){1;2;0}
KV
8 tháng 11 2023
a) Đặt A = \(6^5.5-3^5\)
\(=\left(2.3\right)^5.5-3^5\)
\(=2^5.3^5.5-3^5\)
\(=3^5.\left(2^5.5-1\right)\)
\(=3^5.\left(32.5-1\right)\)
\(=3^5.159\)
\(=3^5.3.53⋮53\)
Vậy \(A⋮53\)
b) Đặt \(B=2+2^2+2^3+...+2^{120}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)
\(=2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{119}.\left(1+2\right)\)
\(=2.3+2^3.3+...+2^{119}.3\)
\(=3.\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(B⋮3\)
\(B=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2\right)+3^4.\left(1+2+2^2\right)+...+2^{118}.\left(1+2+2^2\right)\)
\(=2.7+2^4.7+...+2^{118}.7\)
\(=7.\left(2+2^4+...+2^{118}\right)⋮7\)
Vậy \(B⋮7\)
\(B=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(+...+\left(2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)
\(+2^{116}.\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+2^6.31+...+2^{116}.31\)
\(=31.\left(2+2^6+...+2^{116}\right)⋮31\)
Vậy \(B⋮31\)
\(B=\left(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}+2^{13}+2^{14}+2^{15}+2^{16}\right)\)
\(+...+\left(2^{113}+2^{114}+2^{115}+2^{116}+2^{117}+2^{118}+2^{119}+2^{120}\right)\)
\(=2.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)+2^9.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)
\(+...+2^{113}.\left(1+2+2^2+2^3+2^4+2^5+2^6+2^7\right)\)
\(=2.255+2^9.255+...+2^{113}.255\)
\(=255.\left(2+2^9+...+2^{113}\right)\)
\(=17.15.\left(2+2^9+...+2^{113}\right)⋮17\)
Vậy \(B⋮17\)
KV
8 tháng 11 2023
c) Đặt C = \(3^{4n+1}+2^{4n+1}\)
Ta có:
\(3^{4n+1}=\left(3^4\right)^n.3\)
\(2^{4n}=\left(2^4\right)^n.2\)
\(3^4\equiv1\left(mod10\right)\)
\(\Rightarrow\left(3^4\right)^n\equiv1^n\left(mod10\right)\equiv1\left(mod10\right)\)
\(\Rightarrow3^{4n+1}\equiv\left(3^4\right)^n.3\left(mod10\right)\equiv1.3\left(mod10\right)\equiv3\left(mod10\right)\)
\(\Rightarrow\) Chữ số tận cùng của \(3^{4n+1}\) là \(3\)
\(2^4\equiv6\left(mod10\right)\)
\(\Rightarrow\left(2^4\right)^n\equiv6^n\left(mod10\right)\equiv6\left(mod10\right)\)
\(\Rightarrow2^{4n+1}\equiv\left(2^4\right)^n.2\left(mod10\right)\equiv6.2\left(mod10\right)\equiv2\left(mod10\right)\)
\(\Rightarrow\) Chữ số tận cùng của \(2^{4n+1}\) là \(2\)
\(\Rightarrow\) Chữ số tận cùng của C là 5
\(\Rightarrow C⋮5\)
NV
1
LT
19 tháng 6 2021
a) \(\Leftrightarrow x^2-36=64\)
\(\Leftrightarrow x^2=100\)
\(\Leftrightarrow x=\pm10\)
Vậy \(x=\pm10\)
b) \(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}\)
17 tháng 7 2021
a) Ta có: \(x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
b) Ta có: \(81x^4+4y^4\)
\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)
\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)
\(=\left(9x^2-6xy+2y^2\right)\left(9x^2+6xy+2y^2\right)\)
17 tháng 7 2021
c) Ta có: \(x^5+x+1\)
\(=x^5+x^2-x^2+x-1\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
CT
2
TT
3
10 tháng 6 2021
`a)(x+5)^3=-64`
`(x+5)^3=(-4)^3`
`x+5=-4`
`x=-4-5=-9`
Vậy `x=-9`
`2)(2x-3)^3=8`(9 không được)
`(2x-3)^3=2^3`
`2x-3=2`
`2x=5`
`x=5/2`
Vậy `x=5/2`
QT
0
LT
4
HT
9 tháng 8 2021
A = 3 + 32 + 33 + ...+3100
3A = 32 + 33 + 34 + ...+ 3101
3A - A = ( 32 + 33 + 34 + ...+ 3101 ) - ( 3 + 32 + 33 + ...+3100 )
2A = 3101 - 3
Thay vào 2A + 3 = 3n ta có
3101 - 3 + 3 = 3n
3101 = 3n
=> n = 101
9 tháng 8 2021
A = 3 + 32 + 33 +....+ 3100
\(\Rightarrow\) 3A= 3.(3 + 32 + 33 +....+ 3100)
\(\Rightarrow\) 3A= 32 + 33 + 34 +.....+ 3101
\(\Rightarrow\)3A - A= (32 + 33 + 34 +.....+ 3101) - (3 + 32 + 33 +....+ 3100)
\(\Rightarrow\)2A= 3101 - 3
mà 2A + 3 = 3n
\(\Rightarrow\)3101 - 3 + 3 = 3n
\(\Rightarrow\)3101 = 3n
\(\Rightarrow\)n=101
a, đề k rõ :v
b, (x + 2)4 = 625
=> (x + 2) = (+5)4
=> x + 2 = + 5
=> x = 3 hoặc x = -7
vậy_
a, \(4^{n+2}\) = 64
\(4^{n+2}\) = \(4^3\)
n + 2 = 3
n = 3 -2
n = 1
a, \(\left(x+2\right)^4\) = 625
\(\left(x+2\right)^4\) = \(5^4\)
x +2 = 5
x= 5 - 2
x= 3
Tk mk nha