题目

Problem

Use algebra to obtain the set of values of $x$ for which

\begin{align*} \left|\frac{x^2+3x+10}{x+2}\right|<7-x \end{align*}

(9)

解答

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左边是绝对值，所以要同时满足 $\dfrac{x^2+3x+10}{x+2}<7-x$ 和 $\dfrac{x^2+3x+10}{x+2}>-(7-x)$ 。两边分别解完后取交集。注意 $x=-2$ 不在定义域内。

### 答题过程

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First solve

\begin{align*} \frac{x^2+3x+10}{x+2} &<7-x\\[4mm] \frac{x^2+3x+10-(7-x)(x+2)}{x+2} &<0\\[4mm] \frac{x^2+3x+10-(-x^2+5x+14)}{x+2} &<0\\[4mm] \frac{2x^2-2x-4}{x+2} &<0\\[4mm] \frac{2(x-2)(x+1)}{x+2} &<0 \end{align*}

The critical values are $-2$ , $-1$ and $2$ , so

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

Next solve

\begin{align*} \frac{x^2+3x+10}{x+2} &>-(7-x)\\[4mm] \frac{x^2+3x+10+(7-x)(x+2)}{x+2} &>0\\[4mm] \frac{x^2+3x+10+(-x^2+5x+14)}{x+2} &>0\\[4mm] \frac{8x+24}{x+2} &>0\\[4mm] \frac{x+3}{x+2} &>0 \end{align*}

The critical values are $-3$ and $-2$ , so

\begin{align*} x<-3 \quad \text{or} \quad x>-2 \end{align*}

Taking the intersection of the two results,

\begin{align*} x<-3 \quad \text{or} \quad -1<x<2 \end{align*}