\(a,2x\left(x-5\right)+\left(5-x\right)=0\\ =>2x\left(x-5\right)-\left(x-5\right)=0\\ =>\left(2x-1\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}2x-1=0\\x-5=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};5\right\}\)
\(b,\left(2x-3\right)^2-\left(x+5\right)^2=0\\ =>\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\\ =>\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{8;-\dfrac{2}{3}\right\}\)
a.
2x (x - 5) - (x - 5) . 1 = 0
(2x - 1) (x - 5) = 0
2x - 1 = 0 hoặc x - 5 = 0
x = 1/2 hoặc x = 5
b.
(2x - 3 - x - 5) (2x - 3 + x + 5) = 0
(x - 8) (3x + 2) = 0
x - 8 = 0 hoặc 3x + 2 = 0
x = 8 hoặc x = -2/3
