Đáp án:
Giải thích các bước giải:
a) `1/{x-2}+3/{6-x}=2`
ĐKXĐ: `x \ne 2, x \ne 6`
⇔ `{x-6}/{(x-2).(x-6)}-{3.(x-2)}/{(x-2).(x-6)}=\frac{2.(x^2-8x+12)}{(x-2).(x-6)}`
⇔ `x-6-3x+6=2x^2-16x+24`
⇔ `2x^2-14x+24=0`
⇔ `x^2-7x+12=0`
⇔ `(x-4).(x-3)=0`
⇔ \(\left[ \begin{array}{l}x-4=0\\x-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\ (TM)\\x=3\ (TM)\end{array} \right.\)
Vậy `S={4;3}`
b) `(x^2+2x)^2-2(x^2+2x)-3=0`
Đặt `x^2+2x=t`
⇔ `t^2-2t-3=0`
⇔ `(t+1).(t-3)=0`
⇔ \(\left[ \begin{array}{l}t+1=0\\t-3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}t=-1\\t=3\end{array} \right.\)
`t=-1⇒x^2+2x=-1⇔x^2+2x+1=0⇔(x+1)^2=0` (Loại)
`t=3⇒x^2+2x=3⇔x^2+2x-3=0⇔(x-1).(x+3)=0`
⇔ \(\left[ \begin{array}{l}x-1=0\\x+3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Vậy `S={1;-3}`
