a)
25x2 - 9 = 0
⇔(5x)2 - 32 = 0
⇔(5x - 3).(5x + 3)=0
⇔\(\left[{}\begin{matrix}5x-3=0\\5x+3=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}5x=3\\5x=-3\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=\frac{3}{5}\\x=\frac{-3}{5}\end{matrix}\right.\)
b)
(x-3)2 - 4 = 0
⇔(x - 3)2 - 22 = 0
⇔(x - 3 - 2).(x - 3 + 2) = 0
⇔(x - 5).(x - 1) = 0
⇔\(\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
c)
x2 - 2x = 24
⇔ x2 - 2x - 24 = 0
⇔ x2 - 2x + 1 - 24 -1 =0
⇔ (x - 1)2 - 25 = 0
⇔ (x - 1)2 - 52 = 0
⇔ (x - 1 - 5).(x - 1 + 5) = 0
⇔ (x - 6).(x + 4) = 0
⇔ \(\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
Tìm x
a) \(25x^2-9=0\)
\(\Leftrightarrow25x^2=9\)
\(\Leftrightarrow x^2=\frac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{5}\\x=-\frac{3}{5}\end{matrix}\right.\)
Vậy \(x=\left\{\frac{3}{5};-\frac{3}{5}\right\}\)
b) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy x ={5; 1}
c) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=6\end{matrix}\right.\)
Vậy x ={-4; 6}
