Помогите пожалуйста решить неравенства (на фото).. Created by stilnata. algebra-ru

+ - + ------6----9--------x∈(9,+∞) - - + -----[-1]-------[4]-----x∈[-4/3,4]

Помогите пожалуйста решить неравенства (на фото)...

sirca

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$\sqrt{x-5}\ \textless \ x-7\\x \geq 5\\x\ \textgreater \ 7\\x-5\ \textless \ (x-7)^2\\x-5\ \textless \ x^2-14x+49\\x^2-15x+54\ \textgreater \ 0\\(x-6)(x-9)\ \textgreater \ 0\\x_1=6\\x_2=9$
+ - +
------6----9--------
x∈(9,+∞)

$\sqrt{3x+4} \geq x\\x \geq -4/3\\3x+4 \geq \leq x^2\\x^2-3x-4 \leq 0\\(x+1)(x-4) \leq 0\\x_1=-1\\x_2=4$
- - +
-----[-1]-------[4]-----
x∈[-4/3,4]

2 votes Thanks 2

mukus13

Verified answer

$1)$
$\sqrt{x-5} \ \textless \ x-7$

$\left \{ {{x-5 \geq 0} \atop {x-7\ \textgreater \ 0}} \atop{ (\sqrt{x-5})^2\ \textless \ (x-7)^2 }}}.$

$\left \{ {{x \geq 5} \atop {x\ \textgreater \ 7}} \atop{ {x-5}\ \textless \ x^2-14x+49 }}}.$

$\left \{ {{x \geq 5} \atop {x\ \textgreater \ 7}} \atop{-x^2+15x-54\ \textless \ 0 }}}.$

$\left \{ {{x \geq 5} \atop {x\ \textgreater \ 7}} \atop{x^2-15x+54\ \textgreater \ 0 }}}.$

$D=(-15)^2-4*1*54=225-216=9$
$x_1=9$
$x_2=6$

$\left \{ {{x \geq 5} \atop {x\ \textgreater \ 7}} \atop{(x-6)(x-9)\ \textgreater \ 0 }}}.$

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Ответ: $(9;$ $+$ ∞ $)$

$2)$

$\sqrt{3x+4} \geq x$

получаем совокупность из двух условий:

$1)$
$\left \{ {{x\ \textless \ 0} \atop {3x+4 \geq 0}} \right.$

$\left \{ {{x\ \textless \ 0} \atop {3x \geq -4}} \right.$

$\left \{ {{x\ \textless \ 0} \atop {x \geq -1 \frac{1}{3} }} \right.$

$x$ ∈ $[-1 \frac{1}{3};0)$

$2)$
$\left \{ {{x \geq 0} \atop {3x+4 \geq x^2}} \right.$

$\left \{ {{x \geq 0} \atop {x^2-3x-4 \ \leq 0}} \right.$

$D=(-3)^2-4*1*(-4)=9+16=25$
$x_1=4$
$x_2=-1$

$\left \{ {{x \geq 0} \atop {(x+1)(x-4)\ \leq 0}} \right.$

$x$ ∈ $[0;4]$

объединяем оба случая и получаем $x$ ∈ $[-1 \frac{1}{3};4]$

11 votes Thanks 7

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