Rational Expressions and Rational Equations: Very Difficult Problems with Solutions

Solution:
Since [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are distinct, all denominators are non-zero constants. Consider
[tex]P(x)=\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}+\frac{(x-a)(x-b)}{(c-a)(c-b)}-1[/tex]
This is a polynomial in [tex]x[/tex] of degree at most [tex]2[/tex]. For [tex]x=a[/tex] the first fraction equals [tex]1[/tex], and the other two contain the factor [tex]x-a=0[/tex], so
[tex]P(a)=\frac{(a-b)(a-c)}{(a-b)(a-c)}+0+0-1=1-1=0[/tex]
In the same way, for [tex]x=b[/tex] only the second fraction is non-zero and it equals [tex]1[/tex], and for [tex]x=c[/tex] only the third fraction is non-zero and it equals [tex]1[/tex]. Hence [tex]P(b)=P(c)=0[/tex].
Thus [tex]P[/tex] has three distinct roots [tex]a[/tex], [tex]b[/tex], [tex]c[/tex]. A non-zero polynomial of degree at most [tex]2[/tex] has at most two roots, so [tex]P[/tex] is the zero polynomial, i.e. [tex]P(x)=0[/tex] for every [tex]x[/tex]. This means that the given equality holds for every real number [tex]x[/tex].

Solution:
Put [tex]p=\frac{1}{a}[/tex], [tex]q=\frac{1}{b}[/tex], [tex]r=\frac{1}{c}[/tex]. Then [tex]p+q+r=0[/tex] and [tex]pqr=\frac{1}{abc}[/tex]. First we show that [tex]p^3+q^3+r^3=3pqr[/tex]. Indeed, [tex]r=-(p+q)[/tex], so
[tex]r^3=-(p+q)^3=-p^3-q^3-3pq(p+q)=-p^3-q^3+3pqr\ \Rightarrow\ p^3+q^3+r^3=3pqr[/tex]
Moreover,
[tex]\frac{bc}{a^2}=\frac{abc}{a^3}=abc\,p^3,\qquad \frac{ca}{b^2}=\frac{abc}{b^3}=abc\,q^3,\qquad \frac{ab}{c^2}=\frac{abc}{c^3}=abc\,r^3[/tex]
Therefore
[tex]\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=abc\left(p^3+q^3+r^3\right)=3abc\,pqr=\frac{3abc}{abc}=3[/tex]
which is what we had to prove.

Solution:
Domain: [tex]x\ne -1[/tex]. Put
[tex]A=\frac{x(19-x)}{x+1},\qquad B=x+\frac{19-x}{x+1}=\frac{x^2+19}{x+1}[/tex]
Then for every admissible [tex]x[/tex]
[tex]A+B=\frac{19x-x^2+x^2+19}{x+1}=\frac{19(x+1)}{x+1}=19[/tex]
and the equation says [tex]AB=84[/tex]. Hence [tex]A[/tex] and [tex]B[/tex] are the roots of [tex]z^2-19z+84=0[/tex], i.e. [tex]\left\{A, B\right\}=\left\{7, 12\right\}[/tex]. Since [tex]A=19-B[/tex] holds automatically, it is enough to require [tex]B=7[/tex] or [tex]B=12[/tex].
If [tex]B=7[/tex]:
[tex]x^2+19=7x+7\ \Rightarrow\ x^2-7x+12=0\ \Rightarrow\ x=3\text{ or }x=4[/tex]
If [tex]B=12[/tex]:
[tex]x^2+19=12x+12\ \Rightarrow\ x^2-12x+7=0\ \Rightarrow\ x=6\pm\sqrt{29}[/tex]
None of the four numbers equals [tex]-1[/tex], so all of them are roots of the equation: [tex]x_1=3, \ x_2=4, \ x_{3,4}=6\pm\sqrt{29}[/tex].

Solution:
Domain: [tex]x\ne -2[/tex]. The denominators on the left are positive, because
[tex]x^2+2x+2013=(x+1)^2+2012,\quad x^2+2x+2014=(x+1)^2+2013,\quad x^2+2x+2015=(x+1)^2+2014[/tex]
Hence each fraction on the left is at most [tex]1[/tex]:
[tex]\frac{2012}{(x+1)^2+2012}\le 1,\qquad \frac{2013}{(x+1)^2+2013}\le 1,\qquad \frac{2014}{(x+1)^2+2014}\le 1[/tex]
so the left side is at most [tex]3[/tex], with equality only when [tex](x+1)^2=0[/tex], i.e. [tex]x=-1[/tex]. For the right side put [tex]w=(x+2)^2\gt 0[/tex]:
[tex]\frac{(x+2)^4+(x+2)^2+1}{(x+2)^2}=\frac{w^2+w+1}{w}=w+\frac{1}{w}+1=3+\frac{(w-1)^2}{w}\ge 3[/tex]
with equality only when [tex]w=1[/tex], i.e. [tex]x=-1[/tex] or [tex]x=-3[/tex]. Thus left side [tex]\le 3\le[/tex] right side, so the equation holds only if both sides are equal to [tex]3[/tex]. This requires [tex]x=-1[/tex]; for [tex]x=-3[/tex] the left side is less than [tex]3[/tex].
Check: for [tex]x=-1[/tex] the left side is [tex]1+1+1=3[/tex] and the right side is [tex]\frac{1+1+1}{1}=3[/tex]. The equation has the unique solution [tex]x=-1[/tex].

Solution:
Since [tex]a[/tex], [tex]b[/tex], [tex]c[/tex] are non-zero, we multiply each equality by its denominator:
[tex]ab=bt-1,\qquad bc=ct-1,\qquad ca=at-1[/tex]
Multiply the first equality by [tex]c[/tex] and use the second one:
[tex]abc=bct-c=(ct-1)t-c=ct^2-t-c\ \Rightarrow\ abc+t=c\left(t^2-1\right)[/tex]
Multiply the second equality by [tex]a[/tex] and use the third one:
[tex]abc=cat-a=(at-1)t-a=at^2-t-a\ \Rightarrow\ abc+t=a\left(t^2-1\right)[/tex]
Comparing the two results, we get [tex]c\left(t^2-1\right)=a\left(t^2-1\right)[/tex], i.e. [tex](c-a)\left(t^2-1\right)=0[/tex]. Since [tex]a\ne c[/tex], it follows that [tex]t^2=1[/tex], and then
[tex]abc+t=c\left(t^2-1\right)=0[/tex]
which is what we had to prove.

Solution:
First we show that two of the numbers are opposite. Write the condition as
[tex]\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\ \Rightarrow\ \frac{a+b}{ab}=\frac{c-(a+b+c)}{c(a+b+c)}=-\frac{a+b}{c(a+b+c)}[/tex]
Hence
[tex](a+b)\left(\frac{1}{ab}+\frac{1}{c(a+b+c)}\right)=0\ \Rightarrow\ \frac{(a+b)\left(c(a+b+c)+ab\right)}{abc(a+b+c)}=0[/tex]
Since [tex]c(a+b+c)+ab=c^2+ca+cb+ab=(c+a)(c+b)[/tex], we obtain
[tex](a+b)(b+c)(c+a)=0[/tex]
Both the condition and the claim are symmetric in [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], so we may assume that [tex]a+b=0[/tex], i.e. [tex]b=-a[/tex]. For odd [tex]n[/tex] we have [tex]b^n=(-a)^n=-a^n[/tex], so the left side is
[tex]\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n}-\frac{1}{a^n}+\frac{1}{c^n}=\frac{1}{c^n}[/tex]
On the right, [tex]a^n+b^n+c^n=a^n-a^n+c^n=c^n\ne 0[/tex], so the right side is also equal to [tex]\frac{1}{c^n}[/tex]. Hence the equality holds for every odd natural number [tex]n[/tex].

Solution:
Domain: [tex]x\notin\left\{2, 3, 4, 5\right\}[/tex]. Add [tex]1[/tex] to each fraction. The fractions enter with the signs [tex]+[/tex], [tex]-[/tex], [tex]+[/tex], [tex]-[/tex], so the added units cancel out. Since [tex]\frac{x-7}{x-5}+1=\frac{2(x-6)}{x-5}[/tex] (and similarly for the other fractions), the equation becomes
[tex]2(x-6)\left(\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-3}-\frac{1}{x-2}\right)=x-6[/tex]
One root is [tex]x=6[/tex] (it belongs to the domain). For [tex]x\ne 6[/tex] we divide by [tex]x-6[/tex] and group the fractions:
[tex]2\left(\frac{1}{x-5}-\frac{1}{x-2}\right)+2\left(\frac{1}{x-3}-\frac{1}{x-4}\right)=1\ \Rightarrow\ \frac{6}{x^2-7x+10}-\frac{2}{x^2-7x+12}=1[/tex]
Put [tex]k=x^2-7x+11[/tex]; then [tex]x^2-7x+10=k-1[/tex], [tex]x^2-7x+12=k+1[/tex], and in the domain [tex]k\ne\pm 1[/tex]:
[tex]\frac{6}{k-1}-\frac{2}{k+1}=1\ \Rightarrow\ 6(k+1)-2(k-1)=k^2-1\ \Rightarrow\ k^2-4k-9=0\ \Rightarrow\ k=2\pm\sqrt{13}[/tex]
For [tex]k=2+\sqrt{13}[/tex] we get [tex]x^2-7x+9-\sqrt{13}=0[/tex] with discriminant [tex]13+4\sqrt{13}\gt 0[/tex], so there are two real roots [tex]x=\frac{7\pm\sqrt{13+4\sqrt{13}}}{2}[/tex].
For [tex]k=2-\sqrt{13}[/tex] we get [tex]x^2-7x+9+\sqrt{13}=0[/tex] with discriminant [tex]13-4\sqrt{13}=\sqrt{13}\left(\sqrt{13}-4\right)\lt 0[/tex], so there are no real roots.
For the two new roots the number [tex]x^2-7x=\sqrt{13}-9[/tex] is irrational, while for [tex]x\in\left\{2, 3, 4, 5, 6\right\}[/tex] the number [tex]x^2-7x[/tex] is an integer. Hence these roots belong to the domain and are different from [tex]6[/tex]. Therefore the equation has [tex]3[/tex] real roots.