Ecuații în Mulțimea Numerelor Raționale Pozitive
Este vorba de Ecuații de/reductibile_la Tipul $a\cdot x +b=0,$ unde $a,b \in \mathbb{Q_{+}}, x \in \mathbb{Q_{+}}$. Pratic, la acest nivel analizăm ecuația de forma $a\cdot x =b$. Soluția generală este $x=\frac{b}{a}$, în unele cazuri apare cerința $x \in \mathbb{N}$.
Enunt
Rezolvare, 05.11.2002
CĂTĂNEȚ Florina - clasa a VII-a A
MÂRȘA
En, p 415........................................................................................................................
Fr, p 418.....................................................................................................................
ANSWER CiP
Equality $\Leftrightarrow x_{1}=\cdots=x_{n}=16$
SOLUTION CiP
LEMMA If $0\leqslant x <1$ then
(1) $\frac{1}{1-x}\geqslant 1+4x^{2}$.
Equality if and only if $x=0$ or $x=\frac{1}{2}$.
Proof of Lemma
We perform simple calculation
$\frac{1}{1-x}-1-4x^{2}=\frac{x(1-2x)^{2}}{1-x} \geqslant 0$.
The conclusions result.
$\square $(LEMMA)
Let $f(x)=\frac{1}{8-\sqrt{x}}$ where $0<x<64$. If we write
$f(x)=\frac{1}{8} \cdot \frac{1}{1-\frac{\sqrt{x}}{8}} \geqslant \frac{1}{8} \cdot(1+4(\frac{\sqrt{x}}{8})^{2})$
where I applied the Lemma for $\frac {\sqrt{x}}{8}$ instead of $x$, then it turns out
(2) $f(x) \geqslant \frac{1}{8}+\frac{x}{128}$.
We put in the formula (2), one by one $x=x_{1}, x_{2}, \cdots x_{n}$ and add the obtained results.
So we have
$\sum_{i=1}^{n}f(x_{i})\geqslant \sum_{i=1}^{n}\left ( \frac{1}{8}+\frac{x_{i}}{128} \right )=\frac{n}{8}+ \frac{\sum_{i=1}^{n}x_{i}}{128}=\frac{n}{8}+\frac{16 \cdot n}{128}=\frac{n}{8}+\frac{n}{8}=\frac{n}{4}$
what needed to be shown.
$\blacksquare$
The equal sign occur when in (2) eqal signs occur, i.e. $\frac{\sqrt{x_{i}}}{8}=\frac{1}{2}$, so all $x_{i}=16.$
Raspunsul de confirmare
| lun., 12 oct., 19:52 | ||
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Added April 7, 2021
See their solution, V47n03, pages 159-160 (they give two solutions)
The first solution starts from identity
$\frac{1}{1-t}=1+t^2+t+\frac{t^3}{1-t}$
and for the interval $t\in ( 0,\;1 )$ we have
$t+\frac{t^3}{1-t}\geq 3t^2\;(\Leftrightarrow t(2t-1)^2 \geq 0)$.
N.CiP In fact, it is equivalent to inequality (1).
The second solution uses that for $x \in (0, \; 64)$ we have
$\frac{1}{8-\sqrt{x}} \geq \frac {x+16}{128}\; \Leftrightarrow \sqrt{x}(\sqrt{x}-4)^2 \geq 0.$
N. CiP It is essentially the inequality (2).
= end added=
The statement of the problem
En (p 350)
ANSWER CiP
Solutions are the pairs $(x,y)=(k+1,k)\;\;and\;\; (2k+1,2k-1),\;\;\;k=1,2,\cdots$
SOLUTION CiP
Let $z=x-y$. From
$\frac{y(x+y)}{x-y}=\frac{y(x-y)+2y^{2}}{x-y}=y+\frac{2y^{2}}{z}$
we see that $\frac{y(x+y)}{x-y}\in \mathbb{Z}$ $\Leftrightarrow$ $\frac{2y^{2}}{z}\in \mathbb{Z}$. Hence, if $z=1$ or $z=2$, we get respectively $x-y=1$, $x-y=2$. In case $x-y=1$ we find $y=k,\;x=k+1$; in case $x-y=2$ we must $x,y$-odd numbers (otherwise $gcd(x,y)\neq1$) so $y=2k-1,\;x=2k+1$.
Let now $z=x-y=2^{k}\cdot u$, $u$-odd number.
$\frac{2y^{2}}{z}=\frac{y^{2}}{2^{k-1}\cdot u}$
If $k-1>0$, in order that the latter to be positive integer we must that $2 \mid y^{2}$ and $p \mid y^{2}$, where $p$ denote a prime divisor $p\geq 3$ of $u$. Hence $2 \mid y$ and $p \mid y$ and from $x=(x-y)+y=z+y$ it follow that $2 \mid x$ and $p \mid x$ which contradicts $gcd(x,y)=1$. So $k-1 \leq 0$ and $u=1$ and other solutions no longer exists.
$\blacksquare$
Remark. The solutions are exactly those for which $\frac {x+y}{x-y}$ is a positive integer.
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Added April 7, 2021
They notice more simply that $y$ and $x-y$ are coprime, so in identity $\frac{y(x+y)}{x-y}=y+\frac{2y^2}{x-y}$, $x-y$ must divide $2$, and so is equal $1$, or $2$.
They express the answer
$(x,y)=(u+1,u)\;,\;(v+2,v)$, $u-$positive integer, $v-$positive odd integer.
This coincides with our answer.
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