# Some equations from Russian maths book. [closed]

Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.

$$\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1$$ $$\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1$$ $$\sqrt{ (ax-b)^{3}} - \sqrt{ (b-ax)^{3} } = \frac{65}{8}; a \neq 0$$

• Just curious -- which book is this? – Dejan Govc Jan 8 '16 at 19:00
• I've compared this with the original book. The third equation is misstated. The exponent in the second root is supposed to be $-3$, not $3$. – David Jan 8 '16 at 20:42
• @DejanGovc For the sake of completeness: it's a problem book for high school students wishing to go to a technical college (ed. Skanavi). – TT_ Jan 8 '16 at 21:16
• Why was this even upvoted, let alone on HNQ? It looks like an obvious homework question, and the OP didn't even show much (if any) effort. – March Ho Jan 9 '16 at 4:34
• Transcription of the title from the cover you posted: Сканави М.И. Сборник задач по математике для поступающих во ВТУЗЫ. English transcription: Skanavi M.I. (red.) Sbornik zadach 1 po matematike dlja postupajushchih vo vtuzy. Just in case it helps you if you plan to look for the book @DejanGovc – Martin Sleziak Jan 9 '16 at 14:14

1) $\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)}=1\implies\frac{(x^2-5x+4)(x^2-5x+6)}{(x^2+5x+4)(x^2+5x+6)}=1$

$\implies(x^2-5x+4)(x^2-5x+6)=(x^2+5x+4)(x^2+5x+6)$

We know that the even-indexed terms will cancel out because the LHS is just the RHS with $x\to-x$

$\implies-10x^3-50x=10x^3+50x\implies10x^3+50x=0\implies10x(x^2+5)=0\implies x=0$

2) Note that the denominators of these can be written as $u+2,u-4$ with $u=x^2+3x$. We see:

$\frac{6}{u+2}+\frac{8}{u-4}=1\implies\frac{6}{u+2}-3+\frac{8}{u-4}+2=1-3+2\implies\frac{-3u}{u+2}+\frac{2u}{u-4}=0$

Remove a factor of $u$ and accept $u=0$ as a solution. Then clear fractions:

$\frac{-3}{u+2}+\frac{2}{u-4}=0\implies-3(u-4)+2(u+2)=0\implies16-u=0\implies u=16$

So, $x^2+3x=0, 16$ - we extract $x=0,-3$ for $u=0$ and solve the $u=16$ case with the quadratic formula / your method of choice.

3) Let $(ax-b)^{3/7}=u$, then $u-(-u)=65/8\implies u=65/16$. Nowhere further to go as $a,b$ aren't given - can say $x=(65/16+b)/a$ if you'd like, I suppose.

Perhaps worth noting that $\frac{65}{8}=8+\frac{1}{8}$, as an aside.

• Thank you so much for your anwser. I'm going to analyze it precisely :) – radzak Jan 8 '16 at 19:08
• No worries at all - take your time. With these tricky problems, always keep an eye out for a clever substitution or a symmetry - they tend to be the things that cut a question in half for you. – πr8 Jan 8 '16 at 19:09
• Yes, when solving exercises from this book I always try to keep and eye for it, but I'm not that good at maths. I have recently started to practice more, but I have often problems with these exercises. I attend some maths classes and I have to finish like 10 pages from this book for the next week :) Next pages look frightening to me. On the few pages there are like 5 equations per page, because they are so elaborate. – radzak Jan 8 '16 at 19:18
• No need to multiply numerator/denominator to note the clever symmetry. – vonbrand Jan 8 '16 at 19:47
• In part 1)? I suppose so, though it doesn't seem like it'd be markedly faster/slower to keep it as a fraction. – πr8 Jan 8 '16 at 19:50

For the first one, consider the equation as $$\frac{f(x)}{g(x)}=1$$ However $$f(-x)\equiv g(x)$$

Therefore $f$ and $g$ are reflections of each other in the $y$ axis.

Neither has symmetry in the $y$ axis, and neither have roots of the form $x=\pm a$, so the only intersection is on the $y$ axis, so $x=0$ is the only solution.

• I don't understand this proof. An equation $f(x) = f(-x)$ can have solutions $x \ne 0$, even when $f$ isn't an even function. For example, $f(x) = x^3 - x$ isn't even, but $x = 1$ satisfies $f(x) = f(-x)$. – David Jan 8 '16 at 20:29
• Fair point, so I have edited my answer. Thanks. – David Quinn Jan 8 '16 at 21:00
• I'm don't see why the statement "neither of them have roots of the form $x = \pm a$ is any easier than the original problem. Surely this requires proof. – David Jan 8 '16 at 22:27
• What a clever solution without drowning in details. – user5402 Jan 9 '16 at 12:08

The first equation implies that the product of the distances from $x$ to $-1, -2, -3, -4$ is the same as the product of the distances from $x$ to $1, 2, 3, 4$. (This condition is equivalent to the quotient appearing in the equation being $\pm 1$.) If $x > 0$, the latter distances are smaller than the corresponding former ones, so the latter product is smaller. If $x < 0$, then the former is. So the only possibility is $x = 0$.

In the second equation, make the substitution $u = x^2 + 3x$.

For the third one, you were almost there with what you wrote in the comments. But remember that $\sqrt{-A} = - \sqrt{A}$. You get $(ax - b)^{3/7} = 65/16$. Raise both sides of the equation to the power of $7/3$.

Edit The third equation has a typo in it. It was supposed to read $$\sqrt{ (ax-b)^{3}} - \sqrt{ (b-ax)^{-3} } = \frac{65}{8}.$$ In this case, Write $u = (ax-b)^{3/7}$. Then the equation becomes $u + 1/u = 65/8$. Since after clearing denominators this becomes a quadratic equation, there are at most two possibilities for $u$. Since $u = 8$ and $u = 1/8$ work, these must be the ones. We get $ax - b \in \{128, 1/128\}$, after which it's easy to solve for $x$.

• Very nice, David: +1. How did you get access to the book ? – Georges Elencwajg Jan 9 '16 at 10:49
• @GeorgesElencwajg Thanks. Jatimir gave a link to it in the comments to his question. – David Jan 9 '16 at 15:20
• Thanks for the comment, David. I'm impressed by the difficulty and the sheer number of those problems. I wonder how many technical colleges in the world could ask such exercises at an entrance examination... – Georges Elencwajg Jan 9 '16 at 19:23

First question:

Let $y=(x-2)(x-3)$. So it's equivalent to

$$\frac{y(y-2)}{(y+10x)(y+10x-2)}=1$$ $$y^2-2y=y^2+20xy+100x^2-2y-20x=0$$ $$20x=100x^2+20xy$$

For $x\neq 0$, $$1=5x+(x-2)(x-3)$$ $$x^2=-5$$ $$x=\pm\sqrt{5} i$$

Otherwise $x=0$.

• thank you for your anwser. I understood :) – radzak Jan 8 '16 at 19:30

factorizing the left minus the right-hand side we get $$-\frac{20 x \left(x^2+5\right)}{(x+1) (x+2) (x+3) (x+4)}=0$$ can you proceed from here? making the same with the second equation we get $$-\frac{x (x+3) \left(x^2+3 x-16\right)}{(x-1) (x+1) (x+2) (x+4)}=0$$

• Can you explain how you arrived at the expressions? – Airdish Jan 8 '16 at 18:47
• Thank you for your anwser. In the first equation in numerator I have: $$(x-1)(x-2)(x-3)(x-4)-(x+1)(x+2)(x+3)(x+4)$$ is there any other way than multiplying all the brackets and simplify? – radzak Jan 8 '16 at 18:50