Найдите координату точки С, если АС : СВ=2 : 3, и А(-1), В(9)

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nairahay

10.09.2022

$\sqrt{9x^2-x-10} \geq 3x-2$

Данное неравенство равносильно совокупности двух систем:

$\left \{ {{3x-2\ \textless \ 0} \atop {9x^2-x-10 \geq 0}} \right.$

$\left \{ {{3x\ \textless \ 2} \atop {9(x-1 \frac{1}{9} )(x+1) \geq 0}} \right.$

$\left \{ {{x\ \textless \ \frac{2}{3} } \atop {9(x-1 \frac{1}{9} )(x+1) \geq 0}} \right.$

9x^2-x-10=0

$x_1= \frac{1+19}{18} = \frac{10}{9}=1 \frac{1}{9}$
$x_2= \frac{1-19}{18} = -1$
$9x^2-x-10=9(x-1 \frac{1}{9} )(x+1)$

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+ - +
--------[-1]-------------------[10/9]--------------
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∈

∞

$\left \{ {{3x-2 \geq 0} \atop {( \sqrt{9x^2-x-10})^2 \geq (3x-2)^2}} \right.$

$\left \{ {{3x \geq 2} \atop {9x^2-x-10\geq 9x^2-12x+4}} \right.$

$\left \{ {{x \geq \frac{2}{3} } \atop {9x^2-x-10- 9x^2+12x-4 \geq 0}} \right.$

$\left \{ {{x \geq \frac{2}{3} } \atop {11x \geq 14}} \right.$

$\left \{ {{x \geq \frac{2}{3} } \atop {x \geq 1 \frac{3}{11} }} \right.$

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