Đáp án:
Giải thích các bước giải:
`2\sqrt(2x^2 -3x+5)=2x^2 -3x-6`
ĐK: `2x^2-3x+5 > 0` (lđ)
Đặt `\sqrt{2x^2-3x+5}=t\ (t \ge 0)`
`2t=t^2-11`
`⇔ t^2-2t-11=0`
`⇔` \(\left[ \begin{array}{l}t=1+2\sqrt{3}\ (TM)\\t=1-2\sqrt{3}\ (L)\end{array} \right.\)
`⇒ \sqrt{2x^2-3x+5}=1+2\sqrt{3}`
`⇔ 2x^2-3x+5=13+4\sqrt{3}`
`⇔ 2x^2-3x+8+4\sqrt{3}=0`
`Δ=(-3)^2-4.2.(8+4\sqrt{3})`
`Δ=-55-32\sqrt{3}<0`
Vậy PT vô nghiệm
`\sqrt(2x^2 +3x+9) +2x^2+3x=33`
`⇔ \sqrt{2x^2+3x+9}+2x^2+3x-33=0`
ĐK: `2x^2+3x+9 > 0 ∀x` (lđ)
Đặt `\sqrt{2x^2+3x+9}=t\ (t \ge 0)`
`⇔ t+t^2-42=0`
`⇔ (t-6)(t+7)=0`
`⇔` \(\left[ \begin{array}{l}t=6\ (TM)\\t=-7\ (L)\end{array} \right.\)
`\sqrt{2x^2+3x+9}=6`
`⇔ 2x^2+3x+9=36`
`⇔ 2x^2+3x-27=0`
`⇔ (x-3)(2x+9)=0`
`⇔` \(\left[ \begin{array}{l}x=3\\x=-\dfrac{9}{2}\end{array} \right.\)
Vậy `S={3;- 9/2}`
