Question

Me pueden ayudar con los dos problemas porfa

Answer 1

Hola!

Explicación paso a paso:

∆) Solución 01:

$\frac{ax - 1}{1 - bx} = \frac{a}{b} \\ b(ax - 1) = a(1 - bx) \\ abx - b = a - abx \\ abx + abx = a + b \\ 2abx = a + b \\ x = \frac{a + b}{2ab} \\ x = \frac{a}{2ab} + \frac{b}{2ab} \\ x = \frac{1}{2b} + \frac{1}{2a} \\ x = \frac{1}{2} ( \frac{1}{b} + \frac{1}{b} )$

∆) Solución 02:

$\sqrt{1 + x} + \sqrt{x - 2} = \sqrt{2x + 3} \\ {(\sqrt{1 + x} + \sqrt{x - 2})}^{2} = {(\sqrt{2x + 3} )}^{2} \\ {(\sqrt{1 + x} )}^{2} + 2(\sqrt{1 + x} )(\sqrt{x - 2}) + {(\sqrt{x - 2})}^{2} = 2x + 3 \\ (1 + x) + 2( \sqrt{(1 + x)(x - 2)} + (x - 2) = 2x + 3 \\ 1 + x + 2 \sqrt{x - 2 + {x}^{2} - 2x } \: \: + x - 2 = 2x + 3 \\ 2 \sqrt{ {x}^{2} - x - 2 } = 2x + 3 - 2x + 1 \\ 2 \sqrt{ {x}^{2} - x - 2} = 4 \\ \sqrt{ {x}^{2} - x - 2 } = 2 \\ {x}^{2} - x - 2 = {2}^{2} \\ {x}^{2} - x - 2 = 4 \\ {x}^{2} - x = 6 \\ {x}^{2} - x - 6 = 0\\ x \: \: \: \: \: \: \: \: \: - 3 \: \: \: \: \: \: \\ x \: \: \: \: \: \: \: \: \: \: + 2 \: \: \: \: \: \: \\ \\ (x - 3)(x + 2) = 0 \\ \\ x - 3 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: x + 2 = 0 \\ x = 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = - 2$