Question

Помогите пожалуйста не могу решить

Answer 1

Объяснение:

1)

$x-2-\sqrt{2x^2-9x-2}=0\\\\x-2=\sqrt{2x^2-9x-2}$

ОДЗ:

$x-2\geq 0\\\\x\geq 2$

$(x-2)^2=(\sqrt{2x^2-9x-2})^2 \\\\x^2-4x+4=2x^2-9x-2\\\\x^2-5x-6=0\\\\x^2-6x+x-6=0\\\\x*(x-6)+(x-6)=0\\\\(x-6)*(x+1)=0\\\\x-6=0\\\\x_1=6.\\\\x+1=0\\\\x_2=-1\notin.\ \ \ \ \ \ \Rightarrow\\\\x=6.$

Проверка:

$6-2-\sqrt{2*6^2-9*6-2} =0\\\\4-\sqrt{2*36-54-2}=0\\\\4-\sqrt{72-56}=0\\\\4-\sqrt{16}=0\\\\4-4\equiv0.$

Ответ: x=6.

2)

$y=\sqrt{x^2+9}$

ОДЗ:

$y\in[0;+\infty).$

$y^2=(\sqrt{x^2+9})^2\\\\ y^2=x^2+9\\\\x^2=y^2-9\ \ \ \ \ \ \\\\x=б\sqrt{y^2-3^2} =б\sqrt{(y+3)*(y-3)} \\\\(y+3)*(y-3)\geq 0$

-∞__+__-3__-__3__+__+∞                   ⇒

y∈(-∞;-3]U[3;+∞).

Согласно ОДЗ:

Ответ: y∈[3;+∞).

3)

$(x^2-9)*\sqrt{-15+8x-x^2}=0\\ \displaystyle\\\left \{ {{x^2-9=0} \atop {\sqrt{-15+8x-x^2}=0 }} \right. \ \ \ \ \ \ \left \{ {{x^2-3^2=0} \atop {(\sqrt{-15+8x-x^2})^2=0^2 }} \right. \ \ \ \ \ \ \left \{ {{(x+3)*(x-3)=0} \atop {-15+8x-x^2=0\ |*(-1)}} \right. \\\\\\\left \{ {{x_1=-3\ \ \ \ x_2=3} \atop {x^2-8x+15=0}} \right. \ \ \ \ \ \ \left \{ {{x_1=-3\ \ \ \ x_2=3} \atop {x^2-3x-5x+15=0}} \right.\ \ \ \ \ \ \left \{ {{x_1=-3\ \ \ \ x_2=3} \atop {x*(x-3)-5*(x-3)=0 \right.$

$\displaystyle\\\left \{ {{x_1=-3\ \ \ \ x_2=3} \atop {(x-3)*(x-5)=0}} \right.\ \ \ \ \ \ \left \{ {{x_1=-3\ \ \ \ x_2=3} \atop {x_3=3\ \ \ \ x_4=5}} \right. .\\\\\sum=-3+3+5\\\\\sum=5.$