![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow x-3\inƯ\left(21\right)\)
\(\Leftrightarrow x-3\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{0;2;4;6;10;24\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{2;0;18\right\}\)
c: \(\Leftrightarrow2x-1+4⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{1;0\right\}\)
d: \(\Leftrightarrow x^2+x+3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 17-(2x-11) = 12-3x
2x-11 = 17-12+3x
2x-11 = 5+3x
2x-3x = 5+11
-x = 16
x = -16
b, |2x-3|=7
=>2x-3=7 hoặc 2x-3=-7
=>2x=10 hoặc 2x=-4
=>x=5 hoặc x=-2
c, (2x-1)4 = 16
(2x-1)4 = 24
2x-1=2
2x=3
x=3/2
d, (17-2x)3 = -125
(17-2x)3 = (-5)3
17-2x=-5
2x=17-(-5)
2x=22
x=11
a, => 17-2x+11 = 12-3x
=> 28-2x=12-3x
=> 28=12-3x+2x = 12-x
=> x=12-28 = -16
b, => 2x-3=7 hoặc 2x-3=-7
=> x=5 hoặc x=-2
c, => 2x-1=2 hoặc 2x-1=-2
=> x=3/2 hoặc x=-1.2
d, => 17-2x=-5
=> 2x=17-(-5) = 22
=> x=22:2 = 11
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
( x + 22 ) \(⋮\)( x + 1 )
x + 1 + 21 \(⋮\)( x + 1 )
Mà x + 1 \(⋮\)x + 1 → 21 \(⋮\)x + 1 \(\in\)Ư ( 21 )
( x - 2 ) . ( 2y + 1 ) = 17
Mà 17 là số nguyên tố và bằng 1 . 17
→ Nếu ( x - 2 ) = 1 thì ( 2y + 1 ) = 17
→ Nếu ( 2y + 1 ) = 1 thì ( x - 2 ) = 17
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(-3x+2\right)-\left(5-3x\right)=-3\)
\(\Rightarrow-3x+2-5+3x=-3\)
\(\Rightarrow-3x+3x=-3+5-2\)
\(\Rightarrow0x=0\Rightarrow x\in Z\)
\(3+x-\left(3x-1\right)=6-2x\)
\(\Rightarrow3+x-3x+1=6-2x\)
\(\Rightarrow x-3x+2x=6-1-3\)
\(\Rightarrow0x=2\left(loại\right)\)
\(\left(x-5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{4}{3}\end{cases}}}\)
\(7x\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
\(\left(3x-1\right)2x=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=0\end{cases}}}\)
a) \(2^{x-1}+2^{x+1}+2^{x+2}=104\)
=> \(2^{x-1}+2^x\cdot2+2^x\cdot2^2=104\)
=> \(2^x:2+2^x\cdot\left(2+2^2\right)=104\)
=> \(2^x\cdot\frac{1}{2}+2^x\cdot6=104\)
=> \(2^x\cdot\left(\frac{1}{2}+6\right)=104\Rightarrow2^x=104:\left(\frac{1}{2}+6\right)=104:\frac{13}{2}=16\)
=> \(x=4\)
đăng bài thế nào zay
tôi ko biết đăng bài