Question

Tìm x:

a) ( 3x +1)² = 4

Answer 1

\(\left(3x+1\right)^2=4\\ \Leftrightarrow\left(3x+1\right)^2=\left(\pm2\right)^2\\ \Rightarrow3x+1\in\left\{2;-2\right\}\\ \Rightarrow\left[{}\begin{matrix}3x+1=2\\3x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy...

Answer 2

https://i.imgur.com/APyu4IM.jpg

Answer 3

\(\left(3x+1\right)^2=4\Leftrightarrow3x+1=2\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\)

Answer 4

ta có:

(3x+1)^2=4

<=>(3x+1)^2=(2 hoặc-2)

<=>3x+1 thuộc {2;-2}

<=>3x+1=2

3x+1=-2

<=>x=1/3

x=-1

vậy x= 1/3;x=-1

Answer 5

a) \(\left(3x+1\right)^2=4\)

\(\Rightarrow3x+1=\pm2.\)

\(\Rightarrow\left[{}\begin{matrix}3x+1=2\\3x+1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:3\\x=\left(-3\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{3};-1\right\}.\)

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