Решите уравнение V(6x+1) + V(4x+2) = V(8x) + V(2x+3) V- корень

Интерпритируем: V(6x+1) + V(4x+2) = V(8x) + V(2x+3)                [latex] \sqrt{6x+1} + \sqrt{4x+2} = \sqrt{8x} + \sqrt{2x+3} \ ; [/latex] Находим ОДЗ: [latex] \left\{\begin{array}{l} 6x + 1 \geq 0 \ , \\ 4x + 2 \geq 0 \ , \\ 8x \geq 0 \ , \\ 2x + 3 \geq 0 \ ; \end{array}\right [/latex] [latex] \left\{\begin{array}{l} 6x \geq -1 \ , \\ 4x \geq -2 \ , \\ 8x \geq 0 \ , \\ 2x \geq -3 \ ; \end{array}\right [/latex] [latex] \left\{\begin{array}{l} x \geq -\frac{1}{6} \ , \\\\ x \geq -\frac{1}{2} \ , \\\\ x \geq 0 \ , \\\\ x \geq -1.5 \ ; \end{array}\right [/latex] [latex] x \geq 0 \ ; [/latex] Решаем уравнение. Возводим обе части в квадрат: [latex] ( \sqrt{6x+1} + \sqrt{4x+2} )^2 = ( \sqrt{8x} + \sqrt{2x+3} )^2 \ ; [/latex] [latex] ( \sqrt{6x+1} )^2 + 2 \cdot \sqrt{6x+1} \cdot \sqrt{4x+2} + ( \sqrt{4x+2} )^2 = \\\\ = ( \sqrt{8x} )^2 + 2 \cdot \sqrt{8x}\cdot \sqrt{2x+3} + ( \sqrt{2x+3} )^2 \ ; [/latex] [latex] 6x + 1 + 2 \cdot \sqrt{6x+1} \cdot \sqrt{4x+2} + 4x + 2 = 8x + 2 \cdot \sqrt{8x} \cdot \sqrt{2x+3} + 2x + 3 \ ; [/latex] [latex] 10x + 3 + 2 \cdot \sqrt{6x+1} \cdot \sqrt{4x+2} = 10x + 3 + 2 \cdot \sqrt{8x} \cdot \sqrt{2x+3} \ ; [/latex] [latex] 2 \cdot \sqrt{6x+1} \cdot \sqrt{4x+2} = 2 \cdot \sqrt{8x} \cdot \sqrt{2x+3} \ ; [/latex] [latex] \sqrt{6x+1} \cdot \sqrt{4x+2} = \sqrt{8x} \cdot \sqrt{2x+3} \ ; [/latex] Снова возводим обе части в квадрат: [latex] ( \sqrt{6x+1} \cdot \sqrt{4x+2} )^2 = ( \sqrt{8x} \cdot \sqrt{2x+3} )^2 \ ; [/latex] [latex] ( \sqrt{6x+1} )^2 \cdot ( \sqrt{4x+2} )^2 = ( \sqrt{8x} )^2 \cdot ( \sqrt{2x+3} )^2 \ ; [/latex] [latex] (6x+1)(4x+2) = 8x(2x+3) \ ; [/latex] [latex] 24x^2 + 12x + 4x + 2 = 16x^2 + 24x \ ; [/latex] [latex] 8x^2 - 8x + 2 = 0 \ ; [/latex] [latex] 4x^2 - 4x + 1 = 0 \ ; [/latex] [latex] (2x)^2 - 2 \cdot 2x \cdot 1 + 1^2 = 0 \ ; [/latex] [latex] ( 2x - 1 )^2 = 0 \ ; [/latex] [latex] 2x - 1 = 0 \ ; [/latex] [latex] 2x = 1 \ ; [/latex] [latex] x = \frac{1}{2} = 0.5 > 0 \ [/latex]   –   входит в ОДЗ. О т в е т : x = 0.5 .