Đáp án + Giải thích các bước giải:
`2` , `2x^2 - x = 0`
` x(2x - 1) = 0`
⇒\(\left[ \begin{array}{l}x=0\\2x - 1 =0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=0\\x = \frac{1}{2}\end{array} \right.\)
Vậy `S = {0 ; 1/2}`
`4` , `2x(x - 2) + 3(2 - x) = 0`
` (2x - 3)(x - 2) = 0`
⇒\(\left[ \begin{array}{l}x - 2 =0\\2x-3=0 \end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x = \frac{3}{2}\end{array} \right.\)
Vậy `S = {2 ; 3/2}`
`6`, `x(x - 1) - 5(1 - x) = 0`
` (x + 5)(x - 1) = 0`
⇒\(\left[ \begin{array}{l}x+5=0\\x-1=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=-5\\x=1\end{array} \right.\)
Vậy `S = {-5 ; 1}`
`8` , `3x(x + 3) - 6(3 + x) = 0`
` (3x - 6)(x + 3) = 0`
⇒\(\left[ \begin{array}{l}3x-6=0\\x+3=0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
Vậy `S = {2 ; -3}`
`10` , `(2x + 1)^2 - x(1 + 2x) = 0`
` (2x + 1 - x)(2x + 1) = 0`
` (x + 1)(2x + 1) = 0`
⇒\(\left[ \begin{array}{l}x + 1 = 0\\2x + 1 =0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x =-1\\x = -\frac{1}{2}\end{array} \right.\)
Vậy `S = {-1 ; -1/2}`
