5. 用代数方法求满足下列不等式的 $x$ 的取值集合：

[不接受完全基于图像或数值方法的解答。]

先看右边的正负：$x>1$ 时右边为负数，不等式必然成立；$x=1$ 时两边都是零，不成立。

余下只需考虑 $x<1$。先因式分解，再根据绝对值里面的式子是正还是负，分别去掉绝对值并解不等式。最后把各分支的取值范围合并。

If $x>1$, then $3(1-x)<0$, whereas the left-hand side is non-negative. Hence every $x>1$ satisfies the inequality.

At $x=1$, both sides are zero, so $x=1$ is excluded.

Now consider $x<1$. Let

For $x<1$, we have $x-1<0$. Thus $F(x)\geqslant0$ when $x\leqslant-\dfrac32$, and $F(x)<0$ when $-\dfrac32<x<1$.

Case 1: $x\leqslant-\dfrac32$.

Here $|F(x)|=F(x)$, so

Rearranging and factorising,

Since $x-1<0$, this requires $x+3<0$. Therefore

Case 2: $-\dfrac32<x<1$.

Here $|F(x)|=-F(x)$, so

Since $1-x>0$, this requires $x>0$. Hence this case gives

Combining both cases with $x>1$, the solution set is

也可以用平方，把绝对值不等式化成四次不等式。不过右边不一定是非负数，不能一开始就直接说“两边非负”。

仍先处理 $x>1$ 和 $x=1$；在 $x<1$ 的情况下，右边为正数，才可以把两边平方，再用平方差因式分解。

As in the first method, every $x>1$ is a solution, while $x=1$ is not.

For $x<1$, the right-hand side $3(1-x)$ is positive and the left-hand side is non-negative. Squaring therefore gives the equivalent inequality

Let $F(x)=2x^2+x-3$. By the difference of two squares,

The two factors simplify to

and

Thus the squared inequality is

For $x<1$, the factor $4(x-1)^2$ is strictly positive, so

Hence $x<-3$ or $x>0$. Intersecting with $x<1$ gives

Including the solutions $x>1$, the final solution set is