-Cái này áp dụng hằng đẳng thức số 3 á.
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5+x+2\right)\left(2x-5-x-2\right)=0\)
\(\Leftrightarrow\left(3x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
Vậy...
Bài eassy
\(\left(2x-5\right)^2-\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x-5-x-2\right)\left(2x-5+x+2\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy.....................
Một cách khác dài hơn -.-
\(\left(2x-5\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow4x^2-20x+25-x^2-4x-4=0\)
\(\Leftrightarrow3x^2-24x+21=0\)
\(\Leftrightarrow\left(3x^2-21x\right)-\left(3x-21\right)=0\)
\(\Leftrightarrow x\left(3x-21\right)-\left(3x-21\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-21=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=7\end{matrix}\right.\)
Vậy...........
(2x-5)2 - (x+2)2 = 0
<=> 4x2 + 20x - 25 - x2 - 4x - 4 = 0
<=> 3x2 + 16x - 21 = 0
đến đay tịt rồi , 9-7 ko đk , mk đang suy nghĩ
