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a, \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
b, \(\left(x-1\right)^x+2=\left(x-1\right)^x+4\)
\(\Leftrightarrow\left(x-1\right)^x+2-\left(x-1\right)^x-4=0\)
\(\Leftrightarrow-2\ne0\)=> vô nghiệm
c, \(3^x-1+5.3^{3x}-1=162\)
Đề sai.
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a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
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a)25/x+1 - 7/6 = 1/3 - 0,5
25/x+1 - 7/6 = 1/3 -1/2
25/x+1 - 7/6 = -1/6
25/x+1 = -1/6 + 7/6
25/x+1 = 1
25/x+1 = 25/25
=>25 = x + 1
x = 25 - 1
x = 24
a.Ta có:|2x-1|=2x-1\(\Leftrightarrow\)2x-1\(\ge\)0\(\Leftrightarrow\)x\(\ge\)\(\dfrac{1}{2}\)
|2x-1|=1-2x\(\Leftrightarrow\)2x-1<0\(\Leftrightarrow\)x<\(\dfrac{1}{2}\)
ĐK:\(x\ge\dfrac{1}{2}\)
\(2x-1=2x-1\)
\(\Leftrightarrow2x-1-2x+1=0\)
\(\Leftrightarrow0x=0\)
\(\Rightarrow\)Tập no của PT là S={\(\forall x\)|x\(\ge\dfrac{1}{2}\)}
b.|0,5-3x|=3x-0,5\(\Leftrightarrow\)x<2,5
=0,5-3x\(\Leftrightarrow x\ge2,5\)
ĐK:x<2,5
Gỉai
0,5-3x=3x-0,5
\(\Leftrightarrow\)0,5-3x-3x+0,5=0
\(\Leftrightarrow\)1-6x=0
\(\Leftrightarrow x=\dfrac{1}{6}\)(TMĐKXĐ)
\(\Rightarrow\)tập no của PT là S={\(\dfrac{1}{6}\)}
c.|5x+1-10x|=0,5\(\Leftrightarrow\)|1-5x|=0,5\(\Leftrightarrow x< \dfrac{1}{5}\)
\(\Leftrightarrow\)|1-5x|=-0,5\(\Leftrightarrow\)x\(\ge\dfrac{1}{5}\)
ĐK:\(x< \dfrac{1}{5}\)
Gỉai
1-5x=0,5
\(\Leftrightarrow5x=0,5\)
\(\Leftrightarrow x=0,1\)(loại)
\(\Rightarrow pt\) trên vô nghiệm
d.|x+2|-|x-7|=0
ĐK:x\(\ne\pm2\);x\(\ne\pm7\)
Gỉai
\(\left\{{}\begin{matrix}x+2-x+7=0\\x-2-x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\\-2x-9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-9=0\left(KTMĐKXĐ\right)\\x=-4,5\left(TMĐKXĐ\right)\end{matrix}\right.\)
\(\Rightarrow\)tập no của phương trình là S={-4,5}