Question

Рівняння, що зводиться до квадратних. ​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{x^{2} -11x+18}{x-2} =0\\\\\\\left \{ {{x^{2} -11x+18=0} \atop {x-2\neq 0}} \right. \\\\\\\left \{ {{\left[\begin{array}{ccc}x_{1} =9\\x_{2} =2-neyd\end{array}\right } \atop {x\neq 2}} \right. \\\\\\Otvet: \ 9\\\\\\2)\\\\x^{4} -8x^{2} +7=0\\\\x^{2} =m \ \ ; \ \ m\geq 0\\\\m^{2} -8m+7=0\\\\Teorema \ Vieta:\\\\m_{1} +m_{2} =8\\\\m_{1} \cdot m_{2} =7\\\\m_{1} =1\\\\m_{2} =7\\\\1) \ x^{2} =1\\\\x_{1} =-1 \ \ ; \ \ x_{2} =1$

$\displaystyle\bf\\2) \ x^{2} =7\\\\x_{3} =-\sqrt{7} \ \ ; \ \ x_{4} =\sqrt{7} \\\\\\Otvet: \ -1 \ ; \ 1 \ ; -\sqrt{7} \ ; \ \sqrt{7} \\\\\\1)\\\\\boxed{x^{2} -11x+24=(x-3)(x-8)}\\\\2)\\\\7x^{2} +6x-1=0\\\\D=6^{2} -4\cdot 7\cdot (-1)=36+28=64=8^{2} \\\\\\x_{1}=\frac{-6-8}{14}=-1 \\\\\\x_{2} =\frac{-6+8}{14} =\frac{2}{14} =\frac{1}{7} \\\\\\\boxed{7x^{2} +6x-1=7\Big(x+1\Big)\Big(x-\frac{1}{7} \Big)}$