Question

(2y-1)^2-(y+3)^2=0

 

Answer 1

Lời giải:

$(2y-1)^2-(y+3)^2=0$

$\Leftrightarrow (2y-1-y-3)(2y-1+y+3)=0$

$\Leftrightarrow (y-4)(3y+2)=0$

$\Leftrightarrow y-4=0$ hoặc $3y+2=0$

$\Leftrightarrow y=4$ hoặc $y=\frac{-2}{3}$

Answer 2

`(2y-1)^2 -(y+3)^2=0`

\(\Leftrightarrow\left[2y-1-\left(y+3\right)\right]\left[2y-1+\left(y+3\right)\right]=0\\ \Leftrightarrow\left(2y-1-y-3\right)\left(2y-1+y+3\right)=0\\ \Leftrightarrow\left(y-4\right)\left(3y+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-4=0\\3y+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=4\\3y=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}y=4\\y=-\dfrac{2}{3}\end{matrix}\right.\)

Answer 3

`(2y-1)^2 -(y+3)^2 =0`

`<=>[(2y-1)-(y+3)][(2y-1)+(y+3)]=0`

`<=>(2y-1-y-3)(2y-1+y+3)=0`

`<=>(y-4)(3y+2)=0`

\(< =>\left\{{}\begin{matrix}y-4=0\\3y+2=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}y=4\\y=-\dfrac{2}{3}\end{matrix}\right.\)