Question

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Answer 1

Ответ:

Объяснение:

$x^{2} =t\\t^{2} -t-12=0\\D=1+48\\\sqrt{D} =7\\t_{1} =4\\t_{2} =-3\\x_{1} =16$

$x^{2} +2x=t\\t^{2} -7t-8=0\\D=49+32\\\sqrt{D} =9\\t_{1} = 8\\t_{2} =-1\\x^{2} +2x-8=0\\D=4+32\\\sqrt{D} =6\\x_{1} =2\\x_{2} =-4\\x^{2} +2x+1=0\\D=4-4\\\sqrt{D} =0\\x_{1} =1$

$\frac{x^{2} +x-12}{x-3} \\\left \{ {{x^{2} +x-12=0} \atop {x\neq 3}} \right. \\D=1+48\\\sqrt{D} =7\\x_{1} =3 \\x_{2} =-4$

$\frac{2x^{2} -5x+2-4x^{2} -7x-2}{x-2} \\\frac{-2x^{2} -12x}{x-2} \\\left \{ {{-2x(x+6)} \atop {x\neq 2}} \right. \\\left \{ {{x=-6} \atop {x=0}} \right.$

$\frac{2x^{2} +x-6-x-4x+6}{x(x+2)} \\\frac{2x^{2} -4x}{x(x+2)} \\\left \{ {{2x(x-2)} \atop {x\neq 0 x\neq -2}} \right. \\x=2$