Question

$\sqrt{1 + x \sqrt{2 {x}^{2} + 4x } } + 1 = 2x$
решите уравнение​

Answer 1

Ответ:

2

Объяснение:

$\sqrt{1 +x \sqrt{2 {x}^{2} + 4x } } = 2x - 1 \\ 2x - 1 \geqslant 0 = > x \geqslant \frac{1}{2} \\ 1 + x \sqrt{2 {x}^{2} + 4x} = 4 {x}^{2} - 4x + 1 \\ x \sqrt{2 {x}^{2} + 4x} = 4 {x}^{2} - 4x \\ \sqrt{2 {x}^{2} + 4x} = 4 x - 4 \\ 4x - 4 \geqslant 0 = > x \geqslant 1 \\ 2 {x}^{2} + 4x = 16 {x}^{2} - 32x + 16 \\ 14 {x}^{2} - 36x + 16 = 0 \\ 7 {x}^{2} - 18x + 8 = 0 \\ d = {18}^{2} + 4 \times 7 \times 8 = {2}^{2} (81 -56) = {2}^{2} \times 25 \\ \sqrt{d} = 2 \sqrt{25} =10\\ x_{1} = \frac{18 + 10}{2 \times 7} = \frac{9 + 5 }{7} =2\\ x_{2} = \frac{18 - 10}{2 \times 7} = \frac{4 }{7} < 1$

второй корень не удовлетворяет условию x>=1