Question

[Jennifer Phạm]

1) giải các bpt sau:

a) \(\sqrt{\left(x-1\right)\left(4-x\right)}>x-2\)

b) \(\sqrt{12+x-x^2}\le x-11\)

c) \(\left(x-3\right)\sqrt{x^2-4}\le x^2-9\)

Answer 1

[Minh Nguyệt]

a. \(\sqrt{\left(x-1\right)\left(4-1\right)}>x-2\) ⇔ \(\sqrt{-x^2+5x-4}>x-2\)

ĐK: 1 ≤ x ≤ 4 (1)

BPT ⇔ \(\left[{}\begin{matrix}x-2< 0\\\left\{{}\begin{matrix}x-2>0\\-x^2+5x-4>x^2-4x+4\end{matrix}\right.\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x< 2\\\left\{{}\begin{matrix}x>2\\\frac{9-\sqrt{17}}{4}< x< \frac{9+\sqrt{17}}{4}\end{matrix}\right.\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x< 2\\2< x< \frac{9+\sqrt{17}}{4}\end{matrix}\right.\) (2)

Từ (1), (2) suy ra: \(\left[{}\begin{matrix}1\le x< 2\\2< x< \frac{9+\sqrt{17}}{4}\end{matrix}\right.\) ⇔ x ∈ (1; \(\frac{9+\sqrt{17}}{4}\))\(|\left\{2\right\}\)

b. ĐK: -3 ≤ x ≤ 4 (1)

BPT ⇔ \(\left\{{}\begin{matrix}x-11\ge0\\12+x-x^2\le\left(x-11\right)^2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x\ge11\\\forall x\end{matrix}\right.\) ⇔ x ≥ 11 (2)

Từ (1), (2) suy ra: BPT vô nghiệm

c. ĐK: x ≤ -2, x ≥ 2 (1)

BPT ⇔ (x -3)\(\sqrt{x^2-4}\) ≤ (x - 3)(x + 3)

- Xét x = 3 là nghiệm của BPT (2)

- Xét x≠ 3, BPT ⇔ \(\sqrt{x^2-4}\) ≤ x + 3

⇔ \(\left\{{}\begin{matrix}x+3\ge0\\x^2-4\le\left(x+3\right)^2\end{matrix}\right.\) ⇔ \(\left\{{}\begin{matrix}x\ge-3\\x\ge\frac{-5}{2}\end{matrix}\right.\) ⇔ x ≥ \(\frac{-5}{2}\) (3)

Từ (1), (2), (3) suy ra BPT có nghiệm: x ∈ \([\frac{-5}{2};4]\)