Đáp án:
Giải thích các bước giải:
a) `x^2+\sqrt{x^2-3x+5}=3x+7`
`⇔ x^2-3x-7+\sqrt{x^2-3x+5}=0`
`⇔ x^2-3x+5-12+\sqrt{x^2-3x+5}=0`
Đặt `\sqrt{x^2-3x+5}=t\ (t \ge 0)`
`⇔ t^2+t-12=0`
`⇔ (t-3)(t+4)=0`
`⇔` \(\left[ \begin{array}{l}t-3=0\\t+4=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}t=3\ (TM)\\t=-4\ (L)\end{array} \right.\)
`t=3 ⇔ \sqrt{x^2-3x+5}=3`
`⇔ x^2-3x+5=9`
`⇔ x^2-3x-4=0`
`⇔` \(\left[ \begin{array}{l}x=4\\x=-1\end{array} \right.\)
Vậy `S={-1;4}`
b) `3x^2+21x+18+2\sqrt{x^2+7x+7}=2`
`⇔ 3x^2+21x+16+2\sqrt{x^2+7x+7}=0`
`⇔ 3x^2+21x+21-5+2\sqrt{x^2+7x+7}=0`
`⇔ 3(x^2+7x+7)+2\sqrt{x^2+7x+7}-5=0`
Đặt `\sqrt{x^2+7x+7}=t\ (t \ge 0)`
`3t^2+2t-5=0`
`⇔ (t-1)(3t+5)=0`
`⇔` \(\left[ \begin{array}{l}t-1=0\\3t+5=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}t=1\ (TM)\\t=-\dfrac{5}{3}\ (L)\end{array} \right.\)
`\sqrt{x^2+7x+7}=1`
`⇔ x^2+7x+7=1`
`⇔ x^2+7x+6=0`
`⇔` \(\left[ \begin{array}{l}x=-6\\x=-1\end{array} \right.\)
Vậy `S={-1;-6}`
