Question

a) x^4 + 11x^2 − 12 = 0;
б) 3x−9 / x−1 + x+6 / x+1 = 3.

Answer 1

$1)x^{4}+11x^{2}-12=0\\\\x^{2}=m,m\geq0\\\\m^{2} +11m-12=0\\\\D=11^{2}-4*(-12)=121+48=169=13^{2}\\\\m_{1}=\frac{-11+13}{2}=1\\\\m_{2}=\frac{-11-13}{2}=-12$

$2)\frac{3x-9}{x-1}+\frac{x+6}{x+1}=3\\\\\frac{3x-9}{x-1}+\frac{x+6}{x+1}-3=0\\\\\frac{(3x-9)(x+1)+(x+6)(x-1)-3(x-1)(x+1)}{(x-1)(x+1)}=0\\\\\frac{3x^{2}+3x-9x-9+x^{2}-x+6x-6-3x^{2}+3}{(x-1)(x+1)}=0\\\\\frac{x^{2}-x-12 }{(x-1)(x+1)}=0\\\\\left \{ {{x^{2}-x-12=0 } \atop {x\neq1 \ , \ x\neq-1}} \right.\\\\x^{2}-x-12=0\\\\D=(-1)^{2}-4*(-12)=1+48=49=7^{2}\\\\x_{1}=\frac{1-7}{2}=-3\\\\x_{2}=\frac{1+7}{2}=4\\\\Otvet:\boxed{-3 \ ; \ 4}$

Answer 2

$a) \: {x}^{4} + 11 {x}^{2} - 12 = 0 \\ {x}^{2} = y \\ {y}^{2} + 11y - 12 = 0 \\ y_{1} + y_{2} = - 11 \\ y_{1} \times y_{2} = - 12 \\ y_{1} = - 12 \: \: \: \: \: \: \: \: \: \: \: \: \: \: y_{2} = 1 \\ {x}^{2} = - 12 \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} = 1 \\ x_{1} = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = 1$

$b) \: \frac{3x - 9}{x - 1} + \frac{x + 6}{x + 1} = 3 \\ (x≠ \pm1 )\\ \frac{(3x - 9)(x + 1) + (x + 6)(x - 1)}{(x - 1)(x + 1)} = 3 \\ 3 {x}^{2} - 6x - 9 + {x}^{2} + 5x - 6 = 3 {x}^{2} - 3 \\ {x}^{2} - x - 12 = 0 \\ x_{1} + x_{2} = 1\\ x_{1} \times x_{2} = - 12 \\ x_{1} = - 3 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = 4$