Rational Inequalities: Very Difficult Problems with Solutions

Solution:
Domain: [tex]x\neq -2[/tex], [tex]x\neq 1[/tex]. [tex]x^2+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.
The factor [tex](2x-3)^2[/tex] is positive except at [tex]x=\frac{3}{2}[/tex], where the fraction equals [tex]0[/tex], so [tex]x=\frac{3}{2}[/tex] is a solution. For [tex]x\neq\frac{3}{2}[/tex] we solve
[tex]\frac{(x-2)(x+1)}{(x+2)(x-1)}\geq 0[/tex]
Zeros of the numerator and the denominator: [tex]-2, -1, 1, 2[/tex]. Signs from left to right: [tex]+, -, +, -, +[/tex].
Answer: [tex]x\in(-\infty, -2)\cup[-1, 1)\cup\left\{\frac{3}{2}\right\}\cup[2, +\infty)[/tex].
Note the isolated point [tex]\frac{3}{2}[/tex]: it lies in [tex](1, 2)[/tex], where the fraction is negative, but there the fraction equals [tex]0[/tex].

Solution:
Domain: [tex]x\neq\pm 2[/tex]. Move everything to the left side and bring it to a common denominator:
[tex]\frac{5\left(x^2-4\right)-(x+10)(x+2)-4(x+5)(x-2)+x^2+44}{(x-2)(x+2)}\gt 0[/tex]
Numerator: [tex]5x^2-20-x^2-12x-20-4x^2-12x+40+x^2+44=x^2-24x+44=(x-2)(x-22)[/tex]
[tex]\frac{(x-2)(x-22)}{(x-2)(x+2)}=\frac{x-22}{x+2}\gt 0,\ x\neq 2[/tex]
Answer: [tex]x\in(-\infty, -2)\cup(22, +\infty)[/tex].

Solution:
Domain: [tex]x\neq 0[/tex], [tex]x\neq\pm 1[/tex]. Move everything to the left side and bring it to a common denominator:
[tex]\frac{x(x+1)-8\left(x^2-1\right)+15x(x-1)-2x\left(x^2-1\right)}{2x(x-1)(x+1)}\geq 0[/tex]
Numerator: [tex]-2x^3+8x^2-12x+8=-2\left(x^3-4x^2+6x-4\right)=-2(x-2)\left(x^2-2x+2\right)[/tex]
[tex]x^2-2x+2\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.
[tex]\frac{-2(x-2)}{2x(x-1)(x+1)}\geq 0 \Leftrightarrow \frac{x-2}{x(x-1)(x+1)}\leq 0[/tex]
Zeros of the numerator and the denominator: [tex]-1, 0, 1, 2[/tex]. Signs from left to right: [tex]+, -, +, -, +[/tex].
Answer: [tex]x\in(-1, 0)\cup(1, 2][/tex].

Solution:
Let [tex]t=x^2-4x+5=(x-2)^2+1\geq 1[/tex], then [tex]5x^2-20x+26=5t+1[/tex]:
[tex]5t+1\geq\frac{48}{t} \Leftrightarrow 5t^2+t-48\geq 0[/tex] (we multiplied by [tex]t\gt 0[/tex])
[tex]5t^2+t-48=(5t+16)(t-3)\geq 0[/tex]
Since [tex]5t+16\gt 0[/tex], we get [tex]t\geq 3[/tex].
[tex]x^2-4x+5\geq 3 \Leftrightarrow x^2-4x+2\geq 0 \Leftrightarrow x\leq 2-\sqrt{2}[/tex] or [tex]x\geq 2+\sqrt{2}[/tex]
Answer: [tex]x\in\left(-\infty, 2-\sqrt{2}\right]\cup\left[2+\sqrt{2}, +\infty\right)[/tex].

Solution:
Group the factors so that both products contain [tex]x^2-6x[/tex]:
[tex]x(x-6)\times (x-2)(x-4)=\left(x^2-6x\right)\left(x^2-6x+8\right)[/tex]
Let [tex]t=x^2-6x+4[/tex]: [tex](t-4)(t+4)\geq 9 \Leftrightarrow t^2\geq 25 \Leftrightarrow t\leq -5[/tex] or [tex]t\geq 5[/tex].
[tex]t\geq 5:\ x^2-6x-1\geq 0 \Leftrightarrow x\leq 3-\sqrt{10}[/tex] or [tex]x\geq 3+\sqrt{10}[/tex]
[tex]t\leq -5:\ x^2-6x+9\leq 0 \Leftrightarrow (x-3)^2\leq 0 \Leftrightarrow x=3[/tex]
Answer: [tex]x\in\left(-\infty, 3-\sqrt{10}\right]\cup\{3\}\cup\left[3+\sqrt{10}, +\infty\right)[/tex].

Solution:
[tex]\frac{2x^2-x+2}{x^2+1}=\frac{2\left(x^2+1\right)-x}{x^2+1}=2-\frac{x}{x^2+1}[/tex]
Estimate the fraction [tex]\frac{x}{x^2+1}[/tex]. Since [tex](|x|-1)^2\geq 0[/tex]: [tex]x^2+1\geq 2|x| \Rightarrow -\frac{1}{2}\leq\frac{x}{x^2+1}\leq\frac{1}{2}[/tex]
The bounds are attained: [tex]\frac{1}{2}[/tex] at [tex]x=1[/tex] and [tex]-\frac{1}{2}[/tex] at [tex]x=-1[/tex]. Therefore [tex]\frac{3}{2}\leq\frac{2x^2-x+2}{x^2+1}\leq\frac{5}{2}[/tex].
Answer: [tex]a=\frac{5}{2}[/tex], [tex]b=\frac{3}{2}[/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]-9\left(x^2-x+1\right)\leq 3x^2+px-6\leq 6\left(x^2-x+1\right)[/tex]
Right inequality: [tex]3x^2-(p+6)x+12\geq 0[/tex] holds for every [tex]x[/tex] when [tex]D=(p+6)^2-144\leq 0 \Leftrightarrow -18\leq p\leq 6[/tex].
Left inequality: [tex]12x^2+(p-9)x+3\geq 0[/tex] holds for every [tex]x[/tex] when [tex]D=(p-9)^2-144\leq 0 \Leftrightarrow -3\leq p\leq 21[/tex].
Both conditions must hold: [tex]p\in[-3, 6][/tex].