a. (x - 3)2 - 4 = 0
<=> (x - 3)2 - 22 = 0
<=> (x - 3 + 2)(x - 3 - 2) = 0
<=> (x - 1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
b. x2 - 2x = 24
<=> x2 - 2x - 24 = 0
<=> x2 - 6x + 4x - 24 = 0
<=> x(x - 6) + 4(x - 6) = 0
<=> (x + 4)(x - 6) = 0
<=> \(\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=4\)
TH1:\(x-3=2\text{⇒}x=5\)
TH2:\(x-3=-2\text{⇒}x=1\)
\(a\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2-2^2=0\)
\(\Rightarrow\left(x-3-2\right).\left(x-3+2\right)=0\)
\(\Rightarrow\left(x-5\right).\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{1;5\right\}\)
\(b,x^2-2x=24\)
\(\Rightarrow x^2-2x-24=0\)
\(\Rightarrow x^2+4x-6x-24=0\)
\(\Rightarrow x.\left(x+4\right)-6.\left(x+4\right)\)
\(\Rightarrow\left(x-6\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{-4;6\right\}\)
Hoctot
