题目

Problem

Using algebra, solve the inequality

\begin{align*} \frac{1}{x + 2} > 2x + 3 \end{align*}

(5)

解答

展开

题目要求用代数。分母 $x+2$ 的正负不固定，所以不要直接两边同乘。更稳妥的做法是移到同一边，通分后用临界点做符号分析。

展开

Start with

\begin{align*} \frac{1}{x+2}>2x+3 \end{align*}

Bring all terms to the left-hand side:

\begin{align*} \frac{1}{x+2}-(2x+3)>{}&0\\[4mm] \frac{1-(2x+3)(x+2)}{x+2}>{}&0 \end{align*}

Expand the numerator:

\begin{align*} 1-(2x+3)(x+2) ={}&1-(2x^2+7x+6)\\[4mm] ={}&-2x^2-7x-5\\[4mm] ={}&-(2x+5)(x+1) \end{align*}

\begin{align*} \frac{-(2x+5)(x+1)}{x+2}>{}&0\\[4mm] \frac{(2x+5)(x+1)}{x+2}<{}&0 \end{align*}

The critical values are

\begin{align*} x=-\frac52,\qquad x=-2,\qquad x=-1 \end{align*}

Check the sign of $\frac{(2x+5)(x+1)}{x+2}$ :

\begin{array}{c|cccc} \text{interval} & x<-\frac52 & -\frac52<x<-2 & -2<x<-1 & x>-1\\ \hline \text{sign} & - & + & - & + \end{array}

We need the expression to be negative. Since the original expression is undefined at $x=-2$ , and the inequality is strict, the solution is

\begin{align*} \boxed{x<-\frac52\quad\text{or}\quad -2<x<-1} \end{align*}

展开

因为分母 $x+2$ 的正负号决定了不等号是否需要翻转，所以可以直接在不等式两边同乘分母的平方 $(x+2)^2$ （因为 $x \ne -2$ ，所以 $(x+2)^2$ 恒正）。如此一来可以避免分区间讨论，直接将原分式不等式化简为三次整式不等式，提取公因式后求解。

展开

Since $x \ne -2$ , the term $(x+2)^2$ is strictly positive. Multiply both sides of the inequality by $(x+2)^2$ :

\begin{align*} \frac{1}{x+2} &> 2x+3\\[3mm] (x+2) &> (2x+3)(x+2)^2 \end{align*}

Move all terms to one side:

\begin{align*} (x+2) - (2x+3)(x+2)^2 &> 0 \end{align*}

Factor out the common term $(x+2)$ :

\begin{align*} (x+2)\left[ 1 - (2x+3)(x+2) \right] &> 0\\[4mm] (x+2)\left[ 1 - (2x^2+7x+6) \right] &> 0\\[4mm] (x+2)\left( -2x^2-7x-5 \right) &> 0\\[4mm] -(x+2)(2x+5)(x+1) &> 0\\[4mm] (x+2)(2x+5)(x+1) &< 0 \end{align*}

The critical values are:

\begin{align*} x = -\frac52,\quad x = -2,\quad x = -1 \end{align*}

By testing the intervals (or using a sign chart), the product is negative on:

\begin{align*} \boxed{x<-\frac52\quad\text{or}\quad -2<x<-1} \end{align*}