Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: Ta có: \(x\in B\left(15\right)\)
nên \(x\in\left\{0;15;30;45;60;75;...\right\}\)
mà 40<=x<=70
nên \(x\in\left\{45;60\right\}\)
b: \(2011^2\cdot2011^x=2011^7\)
\(\Leftrightarrow x+2=7\)
hay x=5
Bài 1:
a) Ta có \(\left|x\right|\ge0\) (với mọi \(x\))
Mà \(\left|x\right|\le3\)
\(\Rightarrow0\le\left|x\right|\le3\)
\(\Rightarrow\left|x\right|\in\left\{0;1;2;3\right\}\)
\(\Rightarrow x\in\left\{0;1;2;3;-1;-2;-3\right\}\)
b) Ta có: \(\left|x-1\right|\ge0\) (với mọi \(x\))
Mà \(\left|x-1\right|\le4\)
\(\Rightarrow0\le\left|x-1\right|\le4\)
\(\Rightarrow\left|x-1\right|\in\left\{0;1;2;3;4\right\}\)
\(\Rightarrow x-1\in\left\{0;1;-1;2;-2;3;-3;4;-4\right\}\)
\(\Rightarrow x\in\left\{1;2;0;3;-1;4;-2;5;-3\right\}\)
Bài 2:
\(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow A=2+2+2^2+2^3+2^4+...+2^{20}\)
Đặt \(B=2+2^2+2^3+...+2^{20}\)
\(\Rightarrow2B=2^2+2^3+2^4+...+2^{21}\)
\(\Rightarrow2B-B=\left(2^2+2^3+2^4+...+2^{21}\right)-\left(2+2^2+2^3+...+2^{20}\right)\)
\(\Rightarrow B=2^{21}-2\)
\(\Rightarrow A=2+2^{21}-2\)
\(\Rightarrow A=2^{21}\)
a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
1)
a)-24+3(x-4)=111
3(x-4)=111-(-24)
3(x-4)=111+24
3(x-4)=135
x-4=135:3
x-4=45
x =45+4
x =49
b)(2x-4)(3x+63)=0
\(\Rightarrow\)\(\orbr{\begin{cases}2x-4=0\\3x+63=0\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=2\\x=-21\end{cases}}\)
Vậy x\(\in\){2;-21}
c)|x-7|-4=(-2)4
|x-7| =(-2)4+4
|x-7| =16+4
|x-7| =20
\(\Rightarrow\)\(\orbr{\begin{cases}x-7=7\\x-7=-7\end{cases}}\)\(\Rightarrow\)\(\orbr{\begin{cases}x=14\\x=0\end{cases}}\)
Vậy x\(\in\){14;0}
d)(x-1)2=144
(x-1)2=122
\(\Rightarrow\)x-1=12
x =12+1
x =13
e)(x+7)3=-8
(x+7)3=(-2)3
\(\Rightarrow\)x+7=-2
x =-2-7
x =-9
2)
a)Ta có:
\(3n+12⋮n-3\)
\(\Rightarrow3n-9+21⋮n-3\)
\(\Rightarrow3\left(n-3\right)+21⋮n-3\)
\(\Rightarrow21⋮n-3\)
\(\Rightarrow n-3\inƯ\left(21\right)\)
\(\Rightarrow n-3\in\left\{1;3;7;21\right\}\)
Ta có bảng sau:
| n-3 | 1 | 3 | 7 | 21 |
| n | 4 | 6 | 10 | 24 |
Vậy\(n\in\left\{4;6;10;24\right\}\)
b)Ta có:
\(n+9⋮n-1\)
\(\Rightarrow n-1+10⋮n-1\)
\(\Rightarrow10⋮n-1\)
\(\Rightarrow\)\(n-1\inƯ\left(10\right)\)
\(\Rightarrow n-1\in\left\{1;2;5;10\right\}\)
Ta có bảng sau:
| n-1 | 1 | 2 | 5 | 10 |
| n | 2 | 3 | 6 | 11 |
Vậy \(n\in\left\{2;3;6;11\right\}\)
a) \(4^x+4^{x+1}=80\)
\(\Rightarrow4^x\left(1+4\right)=80\)
\(\Rightarrow4^x=80:5=16\)
\(\Rightarrow x=2\)
b) \(2^x+7\le39\)
\(\Rightarrow2^x\le32\)
\(\Rightarrow2^x\le2^5\)
\(\Rightarrow x\le5.\)
a/ \(4^x+4^{x+1}=80\)
\(\Leftrightarrow4^x\left(1+4\right)=80\)
\(\Leftrightarrow4^x.5=80\)
\(\Leftrightarrow4^x=16\)
\(\Leftrightarrow4^x=4^2\)
\(\Leftrightarrow x=2\left(tm\right)\)
Vậy ...
b/ \(2^x+7\le39\)
\(\Leftrightarrow2^x\le32\)
\(\Leftrightarrow2^x\le2^5\)
\(\Leftrightarrow x\le5\)
\(\Leftrightarrow x\in\left\{0;1;2;3;4;5\right\}\)
Vậy ..