Решить неравенство

Question 1

Решить неравенство $\sqrt{x+8}\ \textless \ x+2$

Question 2

$\sqrt{x+8}\ \textless \ x+2$

$\left \{ {{x+8 \geq 0} \atop {x+2\ \textgreater \ 0}} \atop {( \sqrt{x+8})^2\ \textless \ (x+2)^2 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {x+8\ \textless \ x^2+4x+4 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {-x^2-3x+4\ \textless \ 0 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {x^2+3x-4\ \textgreater \ 0 } \right.$

$x^2+3x-4=0$

$D=3^2-4*1*(-4)=25$

$x_1= \frac{-3+5}{2} =1$

$x_2= \frac{-3-5}{2} =-4$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {(x-1)(x+4)\ \textgreater \ 0 } \right.$

----------[-8]-------------------------------------------
/////////////////////////////////////////////
-----------------------(-2)------------------------------
///////////////////////////////
+ - +
----------------(-4)------------------(1)---------------
////////////////// //////////////////

Ответ: $(1;+$ ∞ $)$

mukus13_zn · Answer 1 · 2018-05-13T06:55:57+0000

$\sqrt{x+8}\ \textless \ x+2$

$\left \{ {{x+8 \geq 0} \atop {x+2\ \textgreater \ 0}} \atop {( \sqrt{x+8})^2\ \textless \ (x+2)^2 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {x+8\ \textless \ x^2+4x+4 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {-x^2-3x+4\ \textless \ 0 } \right.$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {x^2+3x-4\ \textgreater \ 0 } \right.$

$x^2+3x-4=0$

$D=3^2-4*1*(-4)=25$

$x_1= \frac{-3+5}{2} =1$

$x_2= \frac{-3-5}{2} =-4$

$\left \{ {{x \geq -8} \atop {x\ \textgreater \ -2}} \atop {(x-1)(x+4)\ \textgreater \ 0 } \right.$

----------[-8]-------------------------------------------
/////////////////////////////////////////////
-----------------------(-2)------------------------------
///////////////////////////////
+ - +
----------------(-4)------------------(1)---------------
////////////////// //////////////////

Ответ: $(1;+$ ∞ $)$