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\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
\(\left(3x-1\right)^{10}.\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x-1=0\\3x-1^{10}=1\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{2}{3}ho\text{ặc}x=0\end{cases}}\)
Vậy \(x=\frac{1}{3}ho\text{ặc}x=\frac{2}{3}ho\text{ặc}x=0\)
Tham khảo nhé~
TA CÓ:\(\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)
\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{10}\times\left(3x-1\right)^{10}=0\)
\(\Rightarrow\left(3x-1\right)^{10}\times[1-\left(3x-1\right)^{10}]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{2}{3}\end{cases}}}\)
mik nha. cảm ơn nhìu!! ^^
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\(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)
\(\Rightarrow\left(3x-1\right)^{20}-\left(3x-1\right)^{10}=0\)
\(\Rightarrow\left(3x-1\right)^{10}\left[\left(3x-1\right)^{10}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left[{}\begin{matrix}3x-1=1\\3x-1=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)
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\(b.\) \(\left|3x-1\right|=6\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=6\\3x-1=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=6+1\\3x=-6+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=7\\3x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7:3\\x=-5:3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}}\)
Vậy: \(x\in\left\{\frac{7}{3};-\frac{5}{3}\right\}\)
\(a.\) \(\left|3x-1\right|=0\)
\(\Leftrightarrow3x-1=0\)
\(\Leftrightarrow3x=0+1\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=1:3\)
\(\Leftrightarrow x=\frac{1}{3}\)
Vậy: \(x=\frac{1}{3}\)
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a,
\(|3x-1|=0\)
\(\Rightarrow3x-1=0\)
\(3x=1\)
\(x=\frac{1}{3}\)
a,
|3x-1|=0
=>3x-1=0
3x=0+1
3x=1
x=1/3
b,
|3x-1|=6
=>3x-1=6 hoặc 3x-1=-6
Trường hợp 1:
3x-1=6
3x=6+1
3x=7
3x=7/3
Tường hợp 2:
3x-1=-6
3x=-6+1
3x=-5
x=-5/3
c, Hơi dài nên ở bài sau
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=> 3x+1/3 = 0 hoặc x-1/2 = 0
=> x=-1/9 hoặc x=1/2
Vậy x=-1/9 hoặc x=1/2
k mk nha
\(\left(\frac{3x+1}{3}\right).\left(\frac{x-1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{3x+1}{3}=0\\\frac{x-1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}3x+1=0\\x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=1\end{cases}}}\)
\(\left|3x-1\right|+\left|1-3x\right|=6\)
\(\Leftrightarrow\left|3x-1\right|+\left|3x-1\right|=6\)
\(\Leftrightarrow2\left|3x-1\right|=6\)
\(\Leftrightarrow\left|3x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=3\\3x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=4\\3x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)