Reciprocal equation problems: Very Difficult Problems with Solutions

Solution:
[tex]x\ne 0[/tex]. Substitute [tex]y=\dfrac{x}{3}-\dfrac{4}{x}, y^2=\dfrac{x^2}{9}-\dfrac{8}{3}+\dfrac{16}{x^2}[/tex]:
[tex]\dfrac{x^2}{3}+\dfrac{48}{x^2}=3\left(\dfrac{x^2}{9}+\dfrac{16}{x^2}\right)=3y^2+8[/tex]
[tex]3y^2+8=10y \iff 3y^2-10y+8=0 \Rightarrow y=2, y=\dfrac{4}{3}[/tex]
[tex]y=2:\ x^2-6x-12=0 \Rightarrow x=3\pm\sqrt{21}[/tex]
[tex]y=\dfrac{4}{3}:\ x^2-4x-12=0 \Rightarrow x=6, x=-2[/tex]

Solution:
[tex]x=0:\ 0\ne 1[/tex]. Dividing by [tex]x[/tex]; Substitute [tex]t=4x+\dfrac{7}{x}[/tex]:
[tex]\dfrac{4}{t-8}+\dfrac{2}{t-12}=1 \iff 4(t-12)+2(t-8)=(t-8)(t-12) \iff t^2-26t+160=0 \Rightarrow t=10, t=16[/tex]
[tex]t=10:\ 4x^2-10x+7=0, D\lt 0[/tex] — no real roots.
[tex]t=16:\ 4x^2-16x+7=0 \Rightarrow x=\dfrac{1}{2}, x=\dfrac{7}{2}[/tex]

Solution:
Dividing by [tex]x^2\ne 0[/tex] and substituting [tex]t=x+\dfrac{1}{x}[/tex] [tex]\left(x^2+\dfrac{1}{x^2}=t^2-2\right)[/tex]:
[tex]t^2-(m-1)t+m-2=0 \iff (t-1)(t-m+2)=0[/tex]
For real [tex]x[/tex]: [tex]\left|x+\dfrac{1}{x}\right|\ge 2[/tex]. [tex]t=1[/tex] — no real roots.
[tex]t=m-2:\ |m-2|\lt 2 \iff 0\lt m\lt 4[/tex]

Solution:
Dividing by [tex]x^2\ne 0[/tex] and substituting [tex]t=x+\dfrac{1}{x}[/tex] [tex]\left(x^2+\dfrac{1}{x^2}=t^2-2\right)[/tex]: [tex]h(t)=3t^2+2(a-3)t+2a-6=0[/tex]
A value [tex]|t|\gt 2[/tex] gives two distinct real [tex]x[/tex], [tex]t=\pm 2[/tex] gives [tex]x=\pm 1[/tex], and [tex]|t|\lt 2[/tex] gives no real [tex]x[/tex]. So one root [tex]t[/tex] must lie outside [tex][-2, 2][/tex] and the other in [tex](-2, 2)[/tex]:
[tex]h(2)\times h(-2)\lt 0, h(2)=6(a-1), h(-2)=2(9-a) \Rightarrow (a-1)(9-a)\lt 0 \iff a\lt 1[/tex] or [tex]a\gt 9[/tex]
Boundary cases: for [tex]a=1[/tex] we get [tex]t=2[/tex] or [tex]t=-\dfrac{2}{3}[/tex] (only [tex]x=1[/tex]); for [tex]a=9[/tex] the double root [tex]t=-2[/tex] gives only [tex]x=-1[/tex].

Solution:
Dividing by [tex]x^2\ne 0[/tex] and substituting [tex]t=x+\dfrac{1}{x}[/tex] [tex]\left(x^2+\dfrac{1}{x^2}=t^2-2\right)[/tex]: [tex]f(t)=t^2+pt+3p+27=0[/tex]
Four distinct real [tex]x[/tex] require two distinct roots [tex]t[/tex], both with [tex]|t|\gt 2[/tex].
[tex]D=(p-18)(p+6), f(2)=5p+31, f(-2)=p+31[/tex]
[tex]t_1\lt -2\lt 2\lt t_2:\ f(2)\lt 0, f(-2)\lt 0 \Rightarrow p\lt -31[/tex]
[tex]2\lt t_1\lt t_2:\ D\gt 0, -\dfrac{p}{2}\gt 2, f(2)\gt 0 \Rightarrow -\dfrac{31}{5}\lt p\lt -6[/tex]
[tex]t_1\lt t_2\lt -2:\ D\gt 0, -\dfrac{p}{2}\lt -2, f(-2)\gt 0 \Rightarrow p\gt 18[/tex]
Answer: [tex]p\in(-\infty, -31)\cup\left(-\dfrac{31}{5}, -6\right)\cup(18, +\infty)[/tex]