Rational Inequalities: Difficult Problems with Solutions

Solution:
The factors [tex](x-3)^2[/tex] and [tex](x-1)^4[/tex] are positive except at [tex]x=3[/tex] and [tex]x=1[/tex], where the product is [tex]0[/tex] (not a solution of a strict inequality).
The factor [tex](x+2)^3[/tex] has the same sign as [tex]x+2[/tex], [tex](6-x)(4-x)=(x-6)(x-4)[/tex].
It remains to solve [tex](x+2)(x+7)(x-4)(x-6)\lt 0,\ x\neq 1,\ x\neq 3[/tex].
Zeros of the factors: [tex]-7, -2, 4, 6[/tex]. Signs from left to right: [tex]+, -, +, -, +[/tex].
The excluded points [tex]1[/tex] and [tex]3[/tex] lie in [tex](-2, 4)[/tex], which is not part of the answer anyway.
Answer: [tex]x\in(-7, -2)\cup(4, 6)[/tex].

Solution:
Odd powers keep the sign of the base, and [tex](x-1)^4\gt 0[/tex] for [tex]x\neq 1[/tex] (at [tex]x=1[/tex] the product is [tex]0[/tex]). So the inequality is equivalent to
[tex](x+4)(x-2)\lt 0,\ x\neq 1 \Rightarrow -4\lt x\lt 2,\ x\neq 1[/tex]
The integer solutions are [tex]-3, -2, -1, 0[/tex].
[tex]-3-2-1+0=-6[/tex]

Solution:
[tex]x^2+x-6=(x+3)(x-2)[/tex], [tex]x^2-4=(x-2)(x+2)[/tex]
Domain: [tex]x\neq\pm 2[/tex].
[tex]\frac{(x+3)^2(x-2)}{(x-2)(x+2)}=\frac{(x+3)^2}{x+2}\geq 0[/tex] (for [tex]x\neq 2[/tex])
The numerator [tex](x+3)^2[/tex] equals [tex]0[/tex] at [tex]x=-3[/tex] (a solution, since the inequality is non-strict) and is positive otherwise. For [tex]x\neq -3[/tex] we need [tex]x+2\gt 0[/tex], i.e. [tex]x\gt -2[/tex], with [tex]x\neq 2[/tex].
Answer: [tex]x\in\{-3\}\cup(-2, 2)\cup(2, +\infty)[/tex].

Solution:
[tex]\frac{(x-2)(x-3)}{(x-2)(x+4)}\leq 0[/tex], Domain: [tex]x\neq 2[/tex], [tex]x\neq -4[/tex].
[tex]\frac{x-3}{x+4}\leq 0 \Rightarrow -4\lt x\leq 3,\ x\neq 2[/tex]
The integer solutions are [tex]-3, -2, -1, 0, 1, 3[/tex] (the number [tex]2[/tex] is excluded because it makes the denominator zero).
[tex]-3-2-1+0+1+3=-2[/tex]

Solution:
Domain: [tex]x\neq 0[/tex], [tex]x\neq -1[/tex]. Move everything to the left side and bring it to a common denominator:
[tex]\frac{3x-1}{2x}-\frac{3x+2}{2(x+1)}=\frac{(3x-1)(x+1)-x(3x+2)}{2x(x+1)}=\frac{3x^2+2x-1-3x^2-2x}{2x(x+1)}=\frac{-1}{2x(x+1)}[/tex]
[tex]\frac{-1}{2x(x+1)}\gt 0 \Leftrightarrow x(x+1)\lt 0 \Leftrightarrow -1\lt x\lt 0[/tex]
Answer: [tex]x\in(-1, 0)[/tex].

Solution:
[tex]\frac{21x-32}{x^2-9x+18}+2=\frac{21x-32+2x^2-18x+36}{x^2-9x+18}=\frac{2x^2+3x+4}{(x-3)(x-6)}\lt 0[/tex]
[tex]2x^2+3x+4\gt 0[/tex] for every [tex]x[/tex] [tex](D=9-32\lt 0)[/tex].
So the fraction is negative exactly when the denominator is negative: [tex](x-3)(x-6)\lt 0 \Rightarrow 3\lt x\lt 6[/tex].
Answer: [tex]x\in(3, 6)[/tex].

Solution:
[tex]x^2-x+1\gt 0[/tex] for every [tex]x[/tex] (its discriminant is negative and the leading coefficient is positive), so it does not affect the sign. Multiply both sides by the positive denominator:
[tex]5x-1\lt (x-1)\left(x^2-x+1\right)=x^3-2x^2+2x-1[/tex]
[tex]x^3-2x^2-3x\gt 0 \Leftrightarrow x(x-3)(x+1)\gt 0[/tex]
Zeros of the factors: [tex]-1, 0, 3[/tex]. Signs from left to right: [tex]-, +, -, +[/tex].
Answer: [tex]x\in(-1, 0)\cup(3, +\infty)[/tex].

Solution:
[tex]\frac{x^2-1}{x^3-1}=\frac{(x-1)(x+1)}{(x-1)\left(x^2+x+1\right)}=\frac{x+1}{x^2+x+1}[/tex] (for [tex]x\neq 1[/tex])
The denominator [tex]x^2+x+1[/tex] is positive for every [tex]x[/tex], so we can multiply by it without changing the inequality sign:
[tex]\frac{x+1}{x^2+x+1}\lt 1 \Leftrightarrow x+1\lt x^2+x+1[/tex]
The other options come from multiplying by [tex]x^3-1[/tex] (or [tex]x-1[/tex]), whose sign is not known.

Solution:
Domain: [tex]x\neq 0[/tex], [tex]x\neq 1[/tex].
Move everything to the left side and bring it to a common denominator:
[tex]\frac{x(x-1)+2x-6(x-1)}{x(x-1)}=\frac{x^2-5x+6}{x(x-1)}=\frac{(x-2)(x-3)}{x(x-1)}\gt 0[/tex]
Zeros of the numerator and the denominator: [tex]0, 1, 2, 3[/tex]. Signs from left to right: [tex]+, -, +, -, +[/tex].
Answer: [tex]x\in(-\infty, 0)\cup(1, 2)\cup(3, +\infty)[/tex].
If we multiplied by [tex]x(x-1)[/tex] without knowing its sign, we would lose the solutions [tex]x\lt 0[/tex].

Solution:
Domain: [tex]x\neq 0[/tex], [tex]x\neq -3[/tex]. The common denominator is [tex]x(x+3)[/tex]:
[tex]\frac{x^2-(x-3)(x+3)-9(1-x)}{x(x+3)}\lt 0 \Leftrightarrow \frac{x^2-x^2+9-9+9x}{x(x+3)}\lt 0 \Leftrightarrow \frac{9x}{x(x+3)}\lt 0[/tex]
[tex]\frac{9}{x+3}\lt 0 \Rightarrow x\lt -3[/tex] (for [tex]x\neq 0[/tex])
Answer: [tex]x\in(-\infty, -3)[/tex].

Solution:
Let [tex]t=\frac{3-x}{x}[/tex] ([tex]x\neq 0[/tex], [tex]x\neq 3[/tex]), then [tex]\frac{x}{3-x}=\frac{1}{t}[/tex] and
[tex]t+\frac{1}{t}=\frac{t^2+1}{t}\geq 0 \Leftrightarrow t\gt 0[/tex] (the numerator [tex]t^2+1[/tex] is positive).
[tex]\frac{3-x}{x}\gt 0 \Leftrightarrow 0\lt x\lt 3[/tex]
The integer solutions are [tex]1[/tex] and [tex]2[/tex]; their sum is [tex]3[/tex].

Solution:
The number [tex]x=1[/tex] is a root of the cubic polynomial: [tex]1-3-6+8=0[/tex]. Hence
[tex]x^3-3x^2-6x+8=(x-1)\left(x^2-2x-8\right)=(x-1)(x-4)(x+2)[/tex]
[tex](x+2)^2(x-1)(x-4)\leq 0[/tex]
At [tex]x=-2[/tex] the left side is [tex]0[/tex], so it is a solution. For [tex]x\neq -2[/tex] we have [tex](x+2)^2\gt 0[/tex], so [tex](x-1)(x-4)\leq 0[/tex], i.e. [tex]1\leq x\leq 4[/tex].
Answer: [tex]x\in\{-2\}\cup[1, 4][/tex], so the smallest integer solution is [tex]-2[/tex].

Solution:
[tex]x^2+6x+8=(x+2)(x+4)[/tex], [tex]x^2+7x+10=(x+2)(x+5)[/tex]
[tex]\frac{x}{(x+2)(x+4)}-\frac{x}{(x+2)(x+5)}=\frac{x\left[(x+5)-(x+4)\right]}{(x+2)(x+4)(x+5)}=\frac{x}{(x+2)(x+4)(x+5)}\gt 0[/tex]
Zeros of the numerator and the denominator: [tex]-5, -4, -2, 0[/tex]. Signs from left to right: [tex]+, -, +, -, +[/tex].
[tex]x\in(-\infty, -5)\cup(-4, -2)\cup(0, +\infty)[/tex]
The negative integer solutions are [tex]-3[/tex] and [tex]-6, -7, \ldots[/tex]; the largest of them is [tex]-3[/tex].