Question

\((2x-5)^2-(x+2)^2=0\)

Answer 1

\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

 

Answer 2

\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)

=> \(\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)

=>\(\left(x-7\right)\left(3x-3\right)=0\)

=> \(x-7=0=>x=7\)

hoặc \(3x-3=0=>x=1\)

Answer 3

`(2x-5)^2-(x-2)^2=0`

`<=>(2x-5)^2=(x-2)^2`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x-2\\2x-5=-x+2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy `S={3 ;7/3}`.

Answer 4

Câu trước sai đề.

`(2x-5)^2-(x+2)^2=0`

`<=> (2x-5)^2=(x+2)^2`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+2\\2x-5=-x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

`S={7;1}`.