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Inequalities (1 Viewer)

Thread starter mtsmahia
Start date Feb 5, 2010

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mtsmahia

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#1

Solve the inequality 1 (divided by) x < 1 (divided by) (x+1)

thanks

ninetypercent

ninety ninety ninety

#2

$x\neq 0, x\neq -1$
$\frac{1}{x} <\frac{1}{x+1}\\\frac{1}{x}.(x+1)^2 < \frac{1}{x+1}. (x+1)^2\\ \frac{(x+1)^2}{x}< (x+1)\\ x^2. \frac{(x+1)^2}{x}< (x+1).x^2\\ x(x+1)^2 - x^2(x+1)<0\\ x(x+1)(x+1 -x)< 0\\ x(x+1)< 0\\-1<x<0$

M

mtsmahia

Member

#3

ninetypercent said:
$x\neq 0, x\neq -1$
$\frac{1}{x} <\frac{1}{x+1}\\\frac{1}{x}.(x+1)^2 < \frac{1}{x+1}. (x+1)^2\\ \frac{(x+1)^2}{x}< (x+1)\\ x^2. \frac{(x+1)^2}{x}< (x+1).x^2\\ x(x+1)^2 - x^2(x+1)<0\\ x(x+1)(x+1 -x)< 0\\ x(x+1)< 0\\-1<x><0$

thanks

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