\(x^2-25-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2-25\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5-1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
+)TH1: \(x-5=0\Leftrightarrow x=5\)
+)TH2: \(x+4=0\Leftrightarrow x=-4\)
Vậy x-5 hoặc x=-4
\(x^2-25-\left(x-5\right)=0\)
⇔ \(x^2\) -25 -x + 5 = 0
⇔ x\(^2\) -x - 20 = 0
⇔ \(x^2+4x-5x-20=0\)
⇔ \(\left(x^2-5x\right)+\left(4x-20\right)=0\)
⇔ x( x - 5 ) + 4( x - 5 ) = 0
⇔ ( x - 5 ) ( x+ 4 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)
\( C1:\\ {x^2} - 25 - \left( {x - 5} \right) = 0\\ \Leftrightarrow \left( {x - 5} \right)\left( {x + 5} \right) - \left( {x - 5} \right) = 0\\ \Leftrightarrow \left( {x - 5} \right)\left( {x + 5 - 1} \right) = 0\\ \Leftrightarrow \left( {x - 5} \right)\left( {x + 4} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x - 5 = 0\\ x + 4 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 5\\ x = - 4 \end{array} \right.\\ C2:{x^2} - 25 - \left( {x - 5} \right) = 0\\ \Leftrightarrow {x^2} - 25 - x + 5 = 0\\ \Leftrightarrow {x^2} - x - 20 = 0\\ \Leftrightarrow {x^2} + 4x - 5x - 20 = 0\\ \Leftrightarrow x\left( {x + 4} \right) - 5\left( {x + 4} \right) = 0\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} x + 4 = 0\\ x - 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - 4\\ x = 5 \end{array} \right. \)
