本题必须展示所有解题步骤。不接受完全依赖计算器技术的解答。

用代数方法求满足

的 $x$ 的取值范围。

因为 $x^2+1$ 永远为正，可以安全地乘过去，不会改变不等号方向。绝对值小于 $2$ 等价于同时满足小于 $2$ 和大于 $-2$。

Since

we can write

First,

gives

So

Next,

gives

So

Combining the two conditions,

也可以先找临界点，再用数轴判断符号。临界点来自分式等于 $2$ 或 $-2$，这比直接平方后解四次式更清楚。

The critical values occur when

or

Since $x^2+1>0$, these give

so

and

so

Let $F(x)=\Big|\dfrac{x^2+5x-2}{x^2+1}\Big|-2$. Since $x^2+1>0$, there are no excluded real values of $x$. The four zeros of $F$ are

Testing one value in each interval gives

Hence $F(x)<0$ on

分母恒正，可以先乘过去。两边都是非负数，再平方并移项，利用平方差公式分解，便能把四次不等式化为四个一次因子的符号判断。

Since $x^2+1>0$ for every real $x$,

Both sides are non-negative, so squaring preserves the inequality. Thus

Using the difference of two squares,

The four simple roots are $-\dfrac53,0,1,4$. The leading coefficient is positive, so the product has signs

Therefore