Rational Expressions and Rational Equations: Difficult Problems with Solutions

Solution:
The denominators are positive for every real [tex]x[/tex], so the domain is [tex]\mathbb{R}[/tex]. Since [tex]x^2+4\ge 4[/tex] and [tex]x^2+5\ge 5[/tex],
[tex]\frac{4}{x^2+4}\le 1,\qquad \frac{5}{x^2+5}\le 1[/tex]
and in both cases equality holds only for [tex]x=0[/tex]. Hence the left-hand side is at most [tex]2[/tex], and it equals [tex]2[/tex] only when both fractions equal [tex]1[/tex], i.e. only for [tex]x=0[/tex].
The same result is obtained by clearing the denominators:
[tex]4(x^2+5)+5(x^2+4)=2(x^2+4)(x^2+5)\iff 2x^4+9x^2=0\iff x^2(2x^2+9)=0[/tex]
Since [tex]2x^2+9\gt 0[/tex], the only real root is [tex]x=0[/tex]. The equation has exactly one real root.

Solution:
Domains: [tex]F[/tex] is defined for [tex]x\ne 0[/tex], [tex]x\ne -1[/tex]; for [tex]G[/tex] the divisor [tex]\left(\frac{x+2}{2}\right)^2[/tex] must also be non-zero, so in addition [tex]x\ne -2[/tex].
Simplify [tex]F[/tex]: since [tex]\frac{x}{x(x+1)}=\frac{1}{x+1}[/tex],
[tex]1-\frac{1}{x+1}+\frac{1}{x}-\frac{x}{x+1}=1+\frac{1}{x}-\frac{x+1}{x+1}=\frac{1}{x}\Rightarrow F(x)=\frac{4}{x}[/tex]
Simplify [tex]G[/tex]:
[tex]\frac{1}{x(x+1)}+\frac{x+3}{x}=\frac{1+(x+3)(x+1)}{x(x+1)}=\frac{x^2+4x+4}{x(x+1)}=\frac{(x+2)^2}{x(x+1)}[/tex]
[tex]G(x)=\frac{(x+2)^2}{x(x+1)}\times \frac{4}{(x+2)^2}=\frac{4}{x(x+1)},\quad x\ne -2[/tex]
The common points satisfy [tex]\frac{4}{x}=\frac{4}{x(x+1)}[/tex] with [tex]x\notin\{-2, -1, 0\}[/tex]. Multiplying by [tex]x(x+1)\ne 0[/tex] gives [tex]x+1=1[/tex], i.e. [tex]x=0[/tex], which is not admissible.
(If the domains are ignored, clearing the denominators leads to [tex]x(x+2)^2=0[/tex], but both [tex]x=0[/tex] and [tex]x=-2[/tex] lie outside the domain of [tex]G[/tex].) Hence the graphs have no common points, and the set of abscissas is empty: [tex]\varnothing[/tex].

Solution:
Domain: [tex]x\ne 1[/tex] and [tex]x\ne -1[/tex]. For admissible [tex]x[/tex]
[tex]\frac{x+1}{1-x^2}=\frac{x+1}{(1-x)(1+x)}=\frac{1}{1-x}=-\frac{1}{x-1}[/tex]
so the first and the third terms cancel and the equation becomes [tex]x^3-4x^2-4x+1=0[/tex]. Group the terms:
[tex]x^3-4x^2-4x+1=(x^3+1)-4x(x+1)=(x+1)(x^2-x+1)-4x(x+1)=(x+1)(x^2-5x+1)[/tex]
The root [tex]x=-1[/tex] is not in the domain (the fractions are not defined there), so it is extraneous. From [tex]x^2-5x+1=0[/tex] we get
[tex]x_{1,2}=\frac{5\pm\sqrt{21}}{2}[/tex]
These numbers are irrational, hence different from [tex]\pm 1[/tex], so both are roots of the equation.

Solution:
For a natural number [tex]n[/tex]
[tex]f(n)f(n+1)=\frac{2}{2n+1}\times \frac{2}{2n+3}=\frac{4}{(2n+1)(2n+3)}=2\left(\frac{1}{2n+1}-\frac{1}{2n+3}\right)[/tex]
because [tex]\frac{1}{2n+1}-\frac{1}{2n+3}=\frac{2}{(2n+1)(2n+3)}[/tex]. Summing for [tex]n=1,2,\ldots,1002[/tex], all intermediate terms cancel:
[tex]2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots+\frac{1}{2005}-\frac{1}{2007}\right)=2\left(\frac{1}{3}-\frac{1}{2007}\right)=2\times \frac{668}{2007}=\frac{1336}{2007}[/tex]

Solution:
Factor the denominators:
[tex]9x^2-6x+1=(3x-1)^2,\qquad 11-9x^2-30x=-(9x^2+30x-11)=-(3x-1)(3x+11)[/tex]
Domain: [tex]x\ne\frac{1}{3}[/tex] and [tex]x\ne-\frac{11}{3}[/tex]. Put [tex]u=3x-1[/tex] and [tex]v=3x+11[/tex]; both are non-zero on the domain and [tex]v-u=12[/tex]. The equation becomes
[tex]-\frac{2}{uv}=\frac{1}{u^2}-\frac{3}{v^2}[/tex]
Multiplying by [tex]u^2v^2\ne 0[/tex]:
[tex]-2uv=v^2-3u^2\iff 3u^2-2uv-v^2=0\iff (3u+v)(u-v)=0[/tex]
The factor [tex]u-v=-12[/tex] never vanishes, so [tex]3u+v=0[/tex]:
[tex]3(3x-1)+(3x+11)=0\iff 12x+8=0\iff x=-\frac{2}{3}[/tex]
This value belongs to the domain ([tex]u=-3[/tex], [tex]v=9[/tex]). Check: the left side is [tex]\frac{2}{27}[/tex] and the right side is [tex]\frac{1}{9}-\frac{3}{81}=\frac{2}{27}[/tex]. The only root is [tex]x=-\frac{2}{3}[/tex].

Solution:
The discriminants of [tex]4x^2-8x+7[/tex] and [tex]4x^2-10x+7[/tex] are [tex]64-112\lt 0[/tex] and [tex]100-112\lt 0[/tex], so the denominators never vanish and the domain is [tex]\mathbb{R}[/tex]. The number [tex]x=0[/tex] is not a root (the left side is [tex]0\ne 1[/tex]).
For [tex]x\ne 0[/tex] divide the numerator and the denominator of each fraction by [tex]x[/tex]:
[tex]\frac{4}{4x+\frac{7}{x}-8}+\frac{3}{4x+\frac{7}{x}-10}=1[/tex]
Put [tex]t=4x+\frac{7}{x}[/tex] (with [tex]t\ne 8[/tex], [tex]t\ne 10[/tex]):
[tex]\frac{4}{t-8}+\frac{3}{t-10}=1\iff 4(t-10)+3(t-8)=(t-8)(t-10)\iff t^2-25t+144=0\iff t=9\text{ or }t=16[/tex]
For [tex]t=9[/tex]: the equation [tex]4x^2-9x+7=0[/tex] has discriminant [tex]81-112\lt 0[/tex], so it has no real roots.
For [tex]t=16[/tex]:
[tex]4x^2-16x+7=0\Rightarrow x=\frac{16\pm 12}{8}\Rightarrow x_1=\frac{1}{2},\ x_2=\frac{7}{2}[/tex]
Both numbers are non-zero, so both are roots of the equation.

Solution:
The expression requires [tex]a\ne 0[/tex], [tex]a\ne \pm 1[/tex]; none of these numbers satisfies [tex]8a^3=6a+1[/tex] (for [tex]a=0[/tex]: [tex]0\ne 1[/tex]; for [tex]a=1[/tex]: [tex]8\ne 7[/tex]; for [tex]a=-1[/tex]: [tex]-8\ne -5[/tex]), so [tex]A[/tex] is defined. With the common denominator [tex]a(a-1)(a+1)=a^3-a[/tex]:
[tex]A=\frac{a^2}{(a-1)(a+1)}+\frac{1}{a(a+1)}=\frac{a^3+a-1}{a^3-a}[/tex]
From the condition [tex]a^3=\frac{6a+1}{8}[/tex], hence
[tex]a^3+a-1=\frac{6a+1+8a-8}{8}=\frac{7(2a-1)}{8},\qquad a^3-a=\frac{6a+1-8a}{8}=\frac{1-2a}{8}[/tex]
Moreover [tex]2a-1\ne 0[/tex], because [tex]a=\frac{1}{2}[/tex] gives [tex]8a^3=1\ne 4=6a+1[/tex]. Therefore
[tex]A=\frac{7(2a-1)}{1-2a}=-7[/tex]

Solution:
Put [tex]p=a[/tex], [tex]q=2b[/tex], [tex]r=3c[/tex]; these numbers are positive and
[tex]\frac{p}{q+r}=\frac{q}{r+p}=\frac{r}{p+q}[/tex]
Adding [tex]1[/tex] to each of the three equal fractions, we get
[tex]\frac{p+q+r}{q+r}=\frac{p+q+r}{r+p}=\frac{p+q+r}{p+q}[/tex]
Since [tex]p+q+r\gt 0[/tex], the denominators are equal: [tex]q+r=r+p=p+q[/tex], hence [tex]p=q=r[/tex], that is [tex]a=2b=3c[/tex]. Write [tex]a=6s[/tex], [tex]b=3s[/tex], [tex]c=2s[/tex] with [tex]s\gt 0[/tex]. Then
[tex]a+b+c=11s,\qquad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{6s}+\frac{2}{6s}+\frac{3}{6s}=\frac{1}{s}[/tex]
[tex]F=11s\times \frac{1}{s}=11[/tex]

Solution:
Let [tex]s=a+b+c[/tex]; then [tex]s\ne 0[/tex] and [tex]a=s-(b+c)[/tex], hence
[tex]\frac{a^2}{b+c}=\frac{a\left(s-(b+c)\right)}{b+c}=\frac{sa}{b+c}-a[/tex]
In the same way
[tex]\frac{b^2}{c+a}=\frac{sb}{c+a}-b,\qquad \frac{c^2}{a+b}=\frac{sc}{a+b}-c[/tex]
Adding the three equalities and using the condition, we get
[tex]0=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=s\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-(a+b+c)=sA-s=s(A-1)[/tex]
Since [tex]s\ne 0[/tex], we get [tex]A-1=0[/tex], i.e. [tex]A=1[/tex].