Lời giải ..............
\(x^2-25=3x-15\)
\(\Leftrightarrow\left(x^2-25\right)-\left(3x-15\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(x=5\) hoặc \(x=-2\)
\(x^2-25=3x-15\Leftrightarrow x^2-25-3x+15=0\Leftrightarrow\left(x^2-5^2\right)-\left(3x-15\right)=0\Leftrightarrow\left(x+5\right)\left(x-5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Ta có \(x^2-25=3x-15\)
<=> \(x^2-3x-10=0\)
<=> \(\left(x^2-2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)-12,25=0\)
<=> \(\left(x-1,5\right)^2=12,25\)
<=> \(\left[{}\begin{matrix}x=\sqrt{12,25}-1,5\\x=1,5-\sqrt{12,25}\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(x^2-25=3x-15\)
=> (x-5)(x+5) = 3(x-5)
=> (x-5)(x+5)-3(x-5) = 0
=> (x-5)(x+5-3) = 0
=> ( x - 5 ) ( x+ 2 ) = 0
=> x = 5
x = -2
