pt <=> (2x-5-x-2).(2x-5+x+2) = 0
<=> (x-7).(3x-3) = 0
<=> x-7=0 hoặc 3x-3=0
<=> x=7 hoặc x=1
Vậy ............
Tk mk nha
(2x-5)²-(x+2)²=0
(2x-5-x-2)(2x-5+x+2)=0
(x-7)(3x-3)=0
\(\Rightarrow\)x=7 hoặc x=1
\(\Leftrightarrow\left(2x-5+x+2\right).\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-3\right).\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-3=0\\x-7=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Vậy.......
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow2x-5=x+2\)
\(\Leftrightarrow2x-x=2+5\)
\(\Leftrightarrow x=7\)
pt <=> (2x-5-x-2).(2x-5+x+2) = 0
<=> (x-7).(3x-3) = 0
<=> x-7=0 hoặc 3x-3=0
<=> x=7 hoặc x=1
Vậy ............
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
Vì
\(\left(2x-5\right)^2\ge0\forall x\in Z\)
\(\left(x+2\right)^2\ge0\forall x\in Z\)
Nên \(\left(2x-5\right)^2-\left(x+2\right)^2=0\)\(\Leftrightarrow2x-5=0\Rightarrow x=2,5\)
\(x+2=0\Rightarrow x=-2\)