\frac{ x^{2} -5x - 6 }{ x^{2} - 1 } \leq \frac{x - 9}{x - 1} + \frac{2}{x - 3}. Created by nibori07. matematika-ru

\frac{ x^{2} -5x - 6 }{ x^{2} - 1 } \leq \frac{x - 9}{x - 1} + \frac{2}{x

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\frac{ x^{2} -5x - 6 }{ x^{2} - 1 } \leq \frac{x - 9}{x - 1} + \frac{2}{x - 3}

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16oce1 $\frac{ x^{2} -5x - 6 }{ x^{2} - 1 } \leq \frac{x - 9}{x - 1} + \frac{2}{x - 3}$
$\frac{(x+1)(x-6)}{x^2-1}\leq\frac{x-9}{x-1}+\frac2{x-3}$
$\frac{(x+1)(x-6)}{(x-1)(x+1)}\leq\frac{x-9}{x-1}+\frac2{x-3}$
$\frac{x-6}{x-1}\leq\frac{x-9}{x-1}+\frac2{x-3}$
$\frac{x-6}{x-1}\leq\frac{(x-9)(x-3)+2(x-1)}{(x-1)(x-3)}$
$\frac{x-6}{x-1}\leq\frac{x^2-10x+25}{(x-1)(x-3)}$
$\frac{x-6}{x-1}\leq\frac{(x-5)^2}{(x-1)(x-3)}$ x≠1
$x-6\leq\frac{(x-5)^2}{(x-3)}$ x≠3
$(x-6)(x-3)\leq(x-5)^2$
$(x-6)(x-3)-(x-5)^2\leq0$
$x^2-3x-6x+18-x^2+10x-25\leq0$
$x-7\leq0$
$x\leq7$

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nibori07 July 2022 | 0 Ответы

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