kaliprasad said:we have
$\dfrac{1}{x}+ \dfrac{1}{x+2}$
=$ \dfrac{2x+2}{x^2+2x}$
$\gt \dfrac{2x+2}{x^2+2x+1}$
$\gt \dfrac{2(x+1)}{(x+1)^2}$
$\gt \dfrac{2}{x+1}$
hence
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\gt\dfrac{3}{x+2}$
further
$\dfrac{1}{x+1}+ \dfrac{1}{x+2}+ \dfrac{1}{x+3}\lt \dfrac{3}{x+1}$
so to get the upper bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{4}\gt \dfrac{3}{x+2}$
so $10\lt x \lt 11$
so $x\le10\cdots(1)$
so to get the lower bound of x we have
$\dfrac{3}{x+1} \gt \dfrac{1}{2}\gt \dfrac{3}{x+2}$
so $4\lt x\lt 5$
so $x\ge5\cdots(2)$
from (1) and (2) $ 5 \le x \le 10$
A positive integer is a whole number greater than zero.
To find the positive integer solutions to an equation, you can start by setting the equation equal to zero and then factoring it. From there, you can find the factors that make the equation equal to zero and those will be the possible solutions.
Yes, there can be multiple positive integer solutions to an equation. For example, the equation x^2 - 4 = 0 has two positive integer solutions: 2 and -2.
To determine if a positive integer solution is correct, you can plug it back into the original equation and see if it satisfies the equation. If it does, then it is a valid solution.
No, not all equations have positive integer solutions. Some equations may have only negative or irrational solutions, while others may have no real solutions at all.