题目

Problem

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

\begin{align*} \frac{x}{2 - x} \leqslant \frac{x + 3}{x} \end{align*}

(Solutions relying entirely on graphical methods are not acceptable.)

(8)

解答

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这是分式不等式，不能直接乘以 $x(2-x)$ ，因为它的正负会随 $x$ 改变。比较稳妥的方法是先移到同一边，通分成一个分式，再用 critical values 分区间判断符号。

注意原式中分母不能为零，所以

\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

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The inequality is

\begin{align*} \frac{x}{2-x}\leqslant \frac{x+3}{x}, \end{align*}

where

\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

Move all terms to the left:

\begin{align*} \frac{x}{2-x}-\frac{x+3}{x}\leqslant0. \end{align*}

Use the common denominator $x(2-x)$ :

\begin{align*} \frac{x}{2-x}-\frac{x+3}{x} =&\,\frac{x^2-(2-x)(x+3)} {x(2-x)}. \end{align*}

Expand the numerator:

\begin{align*} x^2-(2-x)(x+3) =&\,x^2-\left(2x+6-x^2-3x\right)\\[2mm] =&\,x^2-\left(-x^2-x+6\right)\\[2mm] =&\,2x^2+x-6\\[2mm] =&\,(2x-3)(x+2). \end{align*}

So the inequality becomes

\begin{align*} \frac{(2x-3)(x+2)} {x(2-x)} \leqslant0. \end{align*}

The critical values are

\begin{align*} x=-2,\quad 0,\quad \frac32,\quad 2. \end{align*}

Now test the sign of

\begin{align*} \frac{(2x-3)(x+2)} {x(2-x)} \end{align*}

on each interval:

\begin{array}{c|cccccc} x & (-\infty,-2) & -2 & (-2,0) & (0,\frac32) & (\frac32,2) & (2,\infty)\\ \hline \text{sign} & - & 0 & + & - & + & - \end{array}

Since we need the expression to be less than or equal to zero,

\begin{align*} x\leqslant -2, \qquad 0<x\leqslant \frac32, \qquad x>2. \end{align*}

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也可以乘以一个一定非负的表达式，避免不等号方向问题。这里可以乘以

\begin{align*} x^2(2-x)^2. \end{align*}

因为它在定义域内一定为正，所以不等号方向不变。最后仍然要记得排除 $x=0$ 和 $x=2$ 。

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Start with

\begin{align*} \frac{x}{2-x}\leqslant \frac{x+3}{x}, \end{align*}

where

\begin{align*} x\neq0,\qquad x\neq2. \end{align*}

Multiply both sides by $x^2(2-x)^2$ , which is positive for all allowed $x$ :

\begin{align*} x^3(2-x) \leqslant x(x+3)(2-x)^2. \end{align*}

Bring all terms to one side:

\begin{align*} x^3(2-x)-x(x+3)(2-x)^2 \leqslant0. \end{align*}

Factor out $x(2-x)$ :

\begin{align*} x(2-x) \left[x^2-(x+3)(2-x)\right] \leqslant0. \end{align*}

Now simplify the bracket:

\begin{align*} x^2-(x+3)(2-x) =&\,x^2-\left(-x^2-x+6\right)\\[2mm] =&\,2x^2+x-6\\[2mm] =&\,(2x-3)(x+2). \end{align*}

\begin{align*} x(2-x)(2x-3)(x+2)\leqslant0. \end{align*}

Using the critical values

\begin{align*} -2,\quad 0,\quad \frac32,\quad 2, \end{align*}

and excluding $x=0$ and $x=2$ , the solution is

\begin{align*} x\leqslant -2, \qquad 0<x\leqslant \frac32, \qquad x>2. \end{align*}