Question

не photo math пажожда​

Answer 1

а)

$9 {x}^{2} = 81 \\ {x}^{2} = 9 \\ x_1 = 3 \\ x_2= - 3$

б)

$d = 25 - 16 = 9 \\ x_1= \frac{5 + 3}{4} = 2 \\ x_2= \frac{5 - 3}{4} = 0.5$

в)

$t = {x}^{2} \\ {t}^{2} - 5t + 4 = 0 \\ d = 25 - 16 = 9 \\ t_1= \frac{5 + 3}{2} = 4 \\ t_2= \frac{5 - 3}{2} = 1 \\ x = \sqrt{t} \\ x_1= 2 \\ x_2= - 2 \\ x_3= 1 \\ x_4= - 1$

г)

$x \: не= 5,0\\ \frac{4x + 2x - 10}{x(x - 5)} = 1 \\ 4x + 2x - 10 = {x}^{2} - 5x \\ {x}^{2} - 11x + 10 = 0 \\ d = 121 - 40 = 81 \\ x_1= \frac{11 + 9}{2} = 10 \\ x_2= \frac{11 - 9}{2} = 1$

2)

$x \: не = 1 \\ 4 - {x}^{2} \leqslant 0 \\ - 4 + {x}^{2} \geqslant 0 \\ {x}^{2} - 4 = 0 \\ {x}^{2} = 4 \\ x_{1,2} = + , - 2 \\ x \: принадл( - \infty ; - 2][2 ;+ \infty )$