<=> \(x^2-25=10x+35-2x^2-7x\)
<=> \(3x^2-3x-60=0\)
<=> \(x^2-x-20=0\)
<=> \(\left(x-5\right)\left(x+4\right)=0\)
<=> \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)
Vay \(x\in\left\{-4;5\right\}\)
Chuc ban hoc tot
\(x^2-25=\left(5-x\right)\left(2x+7\right)\)
\(\Leftrightarrow x^2-25=10x-2x^2+35-7x\)
\(\Leftrightarrow3x^2-3x-60=0\)
Ta có \(\Delta=3^2+4.3.60=729,\sqrt{\Delta}=27\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3+27}{6}=5\\x=\frac{3-27}{6}=-4\end{cases}}\)
\(^{x^2-25=\left(5-x\right)\left(2x+7\right)}\)
<=> (x-5)(x + 5) - ( 5 - x )(2x+7) = 0
<=>(x-5)(x + 5) + (x - 5)(2x + 7) =0
<=>(x - 5)(x + 5 + 2x + 7) = 0
<=>(x - 5)(3x + 12) = 0
<=>(x - 5)3(x + 4) = 0
TH1: x - 5 = 0 => x=5
TH2: x+4 = 0 =>x = -4
Vậy PT có tập No={5,-4}
x2 - 25 = (5 - x)(2x + 7)
<=> x2 - 25 = 10x - 2x2 + 35 - 7x
<=> 3x2 - 3x - 60 = 0
<=> 3x2 + 12x - 15x - 60 = 0
<=> 3x(x + 4) - 15(x +4) = 0
<=> 3(x - 5)(x + 4) = 0
<=> x = 5 hoặc x = -4
