Bất phương trình 03

Question

$\sqrt{x+3} - \sqrt{7-x} > \sqrt{2x+8}$

$\sqrt{2x + \sqrt{6x^{2}+1}} > x+1$

score 5 · Answer 1

b. Ta có: $2x+\sqrt{6x^2+1}>2x+\sqrt{4x^2}=2x+2|x|\ge0$
Suy ra: $2x+\sqrt{6x^2+1}>0,\forall x\in\mathbb{R}$
Ta có:
$\sqrt{2x+\sqrt{6x^2+1}}>x+1$
$\Leftrightarrow \left[\begin{array}{l}x+1<0\\\left\{\begin{array}{l}x+1\ge0\\2x+\sqrt{6x^2+1}>(x+1)^2\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x<-1\\\left\{\begin{array}{l}x\ge-1\\\sqrt{6x^2+1}>x^2+1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x<-1\\\left\{\begin{array}{l}x\ge-1\\6x^2+1>x^4+2x^2+1\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x<-1\\\left\{\begin{array}{l}x\ge-1\\x^2(x^2-4)<0\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}x<-1\\\left\{\begin{array}{l}x\ge-1\\-2<x<2\\x\ne0\end{array}\right.\end{array}\right.\Leftrightarrow \left[\begin{array}{l}0<x<2\\x<0\end{array}\right.$

score 5 · Answer 2

Bình chọn tăng 5 Bình chọn giảm

a. Điều kiện: $\left\{\begin{array}{l}x+3\ge0\\7-x\ge0\\2x+8\ge0\end{array}\right.\Leftrightarrow -3\le x\le7$
Bất phương trình tương đương với:
$\sqrt{x+3}>\sqrt{7-x}+\sqrt{2x+8}$
$\Leftrightarrow x+3>7-x+2x+8+2\sqrt{(7-x)(2x+8)}$
$\Leftrightarrow -6>\sqrt{(3-x)(5-2x)}$, vô nghiệm.