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a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x-15)5 = (2x-15)7
(2x-15)5-(2x-15)7 = 0
(2x-15)5-[1-(2x-15)2] = 0
=> (2x-15)5 = 0 hoặc 1-(2x-15)2 = 0
=> 2x-15 = 0 hoặc (2x-15)2 = 1
2x = 0+15 hoặc 2x-15 = 1 hoặc 2x-15 = -1
2x = 15 hoặc 2x = 16 hoặc 2x = 14
x = 15:2 hoặc x = 16:2 hoặc x = 14:2
x = 7,5 hoặc x = 8 hoặc x = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x-1\right)+\frac{3}{15}=\frac{3}{2}\)
\(\Rightarrow2x-1=\frac{3}{2}-\frac{3}{15}=\frac{13}{10}\)
\(\Rightarrow2x=\frac{13}{10}+1=\frac{23}{10}\)
\(\Rightarrow x=\frac{23}{20}\)
b) \(x+\frac{46}{15}=1,5\)
\(\Rightarrow x+\frac{46}{15}=\frac{3}{2}\)
\(\Rightarrow x=\frac{3}{2}-\frac{46}{15}\)
\(\Rightarrow x=\frac{-47}{30}\)
c) \(\left(-2x+1\right)+\frac{3}{15}=\frac{5}{3}\)
\(\Rightarrow-2x+1=\frac{5}{3}-\frac{3}{15}=\frac{22}{15}\)
\(\Rightarrow-2x=\frac{7}{15}\Rightarrow x=\frac{-7}{30}\)
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1) PT \(\Leftrightarrow\dfrac{x+3}{15}=\dfrac{4}{15}\) \(\Rightarrow x+3=4\) \(\Rightarrow x=1\)
Vậy ...
2) Mạnh dạn đoán đề là \(\left(2x-5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy ...
3) PT \(\Rightarrow3x-4-2x+5=3\)
\(\Rightarrow x=2\)
Vậy ...
4) PT \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\\dfrac{1}{2}x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy ...
3) Ta có: \(\left(3x-4\right)-\left(2x-5\right)=3\)
\(\Leftrightarrow3x-4-2x+5=3\)
\(\Leftrightarrow x+1=3\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=\pm1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy.........
Ta có: \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-15-1\right)\left(2x-15+1\right)=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(2x-16\right)\left(2x-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{15}{2};8;7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo :
`(2x - 15)^5 = (2x - 15)^3`
`=> (2x - 15)^5 : (2x - 15)^3 = 1`
`=> (2x - 15)^2 = 1`
`=> (2x - 15)^2 = 1^2`
`=>` $\left[\begin{matrix} 2x-15=1\\ 2x-15=-1\end{matrix}\right.$
`=>` $\left[\begin{matrix} 2x=1 + 15\\ 2x=-1 + 15\end{matrix}\right.$
`=>` $\left[\begin{matrix} 2x=16\\ 2x=14\end{matrix}\right.$
`=>` $\left[\begin{matrix} x=8\\ x=7\end{matrix}\right.$
`=> x in {7;8}`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(=>\left[{}\begin{matrix}2x-15=0\\2x-15=1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}2x=15\\2x=16\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\end{matrix}\right.\)
\(=>x\in\left\{\dfrac{15}{2};8\right\}\)
Ta có :
\(\left(2x-15\right)=\left(2x-15\right)^2\)
\(\Leftrightarrow\)\(\left(2x-15\right).1=\left(2x-15\right)\left(2x-15\right)\)
\(\Leftrightarrow\)\(2x-15=1\)
\(\Leftrightarrow\)\(2x=16\)
\(\Leftrightarrow\)\(x=\frac{16}{2}\)
\(\Leftrightarrow\)\(x=8\)
Vậy \(x=8\)
Chúc bạn học tốt ~
Ta có :
(2x−15)=(2x−15)2
⇔(2x−15).1=(2x−15)(2x−15)
⇔2x−15=1
⇔2x=16
⇔x=162
⇔x=8
Vậy x=8
Chúc bạn học tốt