корень из (х+6)=0,25х+0,25. Created by pupsikpupsik112. algebra-ru

корень из (х+6)=0,25х+0,25...

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$\displaystyle \sqrt{x+6} =0,\! 25x+0,\! 25\\ \\ 4\sqrt{x+6} =x+1\\ \\ \begin{Bmatrix}16(x+6)=(x+1)^2\\ x+6\ge 0\qquad \qquad \quad \\ x+1\ge 0\qquad \qquad \quad \end{matrix} \\ \\ \begin{Bmatrix}x^2+2x-16x+1-96=0\\ x\ge -1>-6\qquad \qquad \qquad \\ x\ge -6\qquad \qquad \qquad \qquad \quad \end{matrix}$

$\begin{Bmatrix}x(x-19)+5(x-19)=0\\ x\ge -1\qquad \qquad \qquad \qquad \end{matrix} \\ \\ \begin{Bmatrix}(x-19)(x+5)=0\\ x\ge -1\quad \qquad \qquad \end{matrix} \\ \\ \begin{Bmatrix}x=\{-5;19\}\\ x\ge -1\qquad \end{matrix} \qquad x=19$

Ответ: 19.

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Ответ:

x ∈ {19}

Объяснение:

$\displaystyle \sqrt{x+6} =0,25*x+0,25$ , ОДЗ: x+6≥0 ⇔

⇔ $\displaystyle \left \{ {{x+6\geq 0} \atop {\sqrt{x+6} =0,25*x+0,25}} \right.$ ⇔ $\displaystyle \left \{ {{x\geq -6} \atop {4*\sqrt{x+6} =x+1}} \right.$ ⇔

⇔ $\displaystyle \left \{ {{x\geq -6, x+1\geq 0} \atop {16*(x+6) =(x+1)^2}} \right.$ ⇔ $\displaystyle \left \{ {{x\geq -6, x\geq -1} \atop {16*x+96 =x^2+2*x+1}} \right.$ ⇔

⇔ $\displaystyle \left \{ {{x\geq -1} \atop {x^2-14*x-95=0}} \right.$ ⇔ $\displaystyle \left \{ {{x\geq -1} \atop {(x+5)*(x-19)=0}} \right.$ ⇔

⇔ $\displaystyle \left \{ {{x\geq -1} \atop {x=-5, x=19}} \right.$ ⇔ x=19.

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