a) 4x^2 - 25 +9 + 12x=0
<=> 4x^2 + 12x - 16 = 0
<=> x^2 - 16 + 3x + 12 = 0
<=> (x - 4) (x + 4) + 3(x + 4)= 0
<=> (x + 4) ( x - 1) = 0
<=> x = -4 hoặc x = 1
b) 16 - 25 + k^2 -8k=0
<=> k^2 - 8k - 9 = 0
<=> k^2 - k + 9k - 9 = 0
<=> k (k -1) + 9 (k -1) = 0
<=> (k -1) ( k +9) = 0
<=> k = 1 hoặc k = -9
a, 4x2-25+9+12x=0
⇔4x2+12x-16=0
⇔(4x2-4x )-( 16x-16)=0
⇔4x(x-1)-16(x-1 )=0
⇔(x-1 )( 4x-16)=0
⇔4(x-1 )(x-4 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy \(S=\left\{1;4\right\}\)
b,16-25+k2-8k=0
⇔k2-8k-9=0
⇔(k2+k )-(9k+9 )=0
⇔k(k+1 )-9(k+1 )=0
⇔(k-9 )(k+1 )=0
\(\Leftrightarrow\left\{{}\begin{matrix}k-9=0\\k+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}k=9\\k=-1\end{matrix}\right.\)
Vậy \(S=\left\{-1;9\right\}\)