[latex]\frac{x^3}{2 \sqrt{x-3} } = \frac{3x^{2}+8x }{ \sqrt{x-3} } [/latex]

Решение x³/2√(x - 3) = (3x² + 8x)/√(x - 3) x³/2√(x - 3) - (3x² + 8x)/√(x - 3) = 0 (x³ - 6x²- 16x)/2√(x - 3) = 0 x³ - 6x² - 16x = 0 x - 3 ≠ 0, x ≠ 3 x(x² - 6x - 16) = 0 x₁ = 0 x² - 6x - 16 = 0 x₂ = - 2 x₃ = 8 Ответ:x₁ = 0 ; x₂ = - 2 ; x₃ = 8

[latex] \frac{x^3}{2 \sqrt{x-3}}= \frac{3x^2+8x}{\sqrt{x-3}}\\\\ \frac{x^3}{2 \sqrt{x-3}}- \frac{3x^2+8x}{ \sqrt{x-3} }=0\\\\ \frac{x^3-6x^2-16x}{2\sqrt{x-3}}=0 [/latex] ОДЗ: [latex]x-3\ \textgreater \ 0\\ x\ \textgreater \ 3[/latex] [latex]x^3-6x^2-16x=0\\\\ x(x^2-6x-16)=0\\\\ x=0\\\\ x^2-6x-16=0\\ D=36+64=100; \ \sqrt{D}=10\\\\ x_{1/2}= \frac{6\pm10}{2}\\\\ x_1= \frac{-4}{2}=-2\\\\ x_2=16:2=8\\\\ x_2=8 [/latex] Ответ: [latex]x=8[/latex]