a,(x-5)(x-2)=2(x-2)
⇔x²-2x-5x+10=2x-4
⇔x²-7x-2x=-10-4
⇔x²-9x=-14
⇔x²-9x+14=0
⇔x²-7x-2x+14=0
⇔(x²-7x)-(2x-14)=0
⇔x(x-7)-2x(x-7)=0
⇔(x-7)(x-2x)=0
⇔\(\left[ \begin{array}{l}x-7=0\\x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=7\\x=0\end{array} \right.\)
vậy S={7;0}
b,(x+3)²-25=0
⇔(x+3-5)(x+3+5)=0
⇔\(\left[ \begin{array}{l}x-2=0\\x+8=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2\\x=-8\end{array} \right.\)
vậy S={2 ;-8}
c,(2x-3)²≤(2x+5)(2x-5)
⇔4x²-12x+9≤4x²-25
⇔4x²-4x²-12x≤-9-25
⇔-12x≤-34
⇔x≥$\frac{17}{6}$
vậy S={x║x≥$\frac{17}{6}$}
cho mik ctlhn đi
