chỉ hộ ạ cảm ơn mn nhìu

$\text{Bài 2:}$

$a) (x-6)(x^2-4)=0$

 \(\left[ \begin{array}{l}x-6=0\\x^2-4=0\end{array} \right.\) $\to$\(\left[ \begin{array}{l}x=6\\x=\pm 2\end{array} \right.\)

$\to \text{Vậy S={6; ±2}}$

$b) (2x+5)(4x^2-9)=0$

\(\left[ \begin{array}{l}2x+5=0\\4x^2-9=0\end{array} \right.\) $\to$\(\left[ \begin{array}{l}x=-\dfrac{5}{2}\\x=\pm \dfrac{3}{2}\end{array} \right.\)

$\text{Vậy S}=\pm \dfrac{3}{2}; -\dfrac{5}{2}.$

$c) (x-2)^2(x-9)=0$

\(\left[ \begin{array}{l}x-9=0\\(x-2)^2=0\end{array} \right.\)$\to$ \(\left[ \begin{array}{l}x-2=0\\x=9\end{array} \right.\) $\to \text{Vậy S={2;9}} $

$d) x^2=2x$

$\to x^2-2x=0$

$\to x(x-2)=0$

\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\) $\to \text{ Vậy S={0;2}}$

$e) x^2-2x+1=4$

$\to (x-1)^2-4=0$

$\to (x-1-2)(x-1+2)=0$

$\to (x-3)(x+1)=0$

\(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\) $\to \text{Vậy S={3;-1}}$

$f) (x^2+1)(x-1)=0$

\(\left[ \begin{array}{l}x^2+1=0\\x-1=0\end{array} \right.\) $\to$ \(\left[ \begin{array}{l}x=1\end{array} \right.\) $\to \text{Vậy S={1}}$

$g) 4x^2+4x+1=0$

$\to (2x+1)^2=0$

$\to x=-\dfrac{1}{2}$

$\text{Vậy S=} -\dfrac{1}{2}$

$h) x^2-5x+6=0$

$\to x^2-2x-3x+6=0$

$\to x(x-2)-3(x-2)=0$

$\to (x-2)(x-3)=0$

$\to$ \(\left[ \begin{array}{l}x=2\\x=3\end{array} \right.\) $\to \text{Vậy S={2;3}}$

$i) 2x^2+3x+1=0$

$\to 2x^2+2x+x+1=0$

$\to 2x(x+1)+(x+1)=0$

$\to (x+1)(2x+1)=0$

\(\left[ \begin{array}{l}x+1\\2x+1\end{array} \right.\)$\to$  \(\left[ \begin{array}{l}x=-1\\x=-\dfrac{1}{2}\end{array} \right.\)

$\to \text{Vậy S=} -1;- \dfrac{1}{2}$