Question

Решить уравнение! 7 номер

Answer 1

$\displaystyle\bf\\A_{x} ^{3} =\frac{1}{20} A_{x} ^{4} \\\\\\\frac{x!}{(x-3)!} =\frac{1}{20} \cdot\frac{x!}{(x-4)!} \\\\\\\frac{(x-3)!\cdot(x-2)\cdot(x-1)\cdot x}{(x-3)!} =\frac{1}{20} \cdot\frac{(x-4)!\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x}{(x-4)!} \\\\\\20\cdot(x-2)\cdot(x-1)\cdot x=(x-3)\cdot(x-2)\cdot(x-1)\cdot x\\\\\\(x-2)\cdot(x-1)\cdot x\cdot \Big[20-x+3\Big]=0\\\\\\(x-2)\cdot(x-1)\cdot x\cdot(23-x)=0\\\\\\x-2=0 \ \ \ \Rightarrow \ \ \ x_{1} =2$

$\displaystyle\bf\\x-1=0 \ \ \ \Rightarrow \ \ \ x_{2} =1\\\\\\x_{3} =0\\\\\\23-x=0 \ \ \ \Rightarrow \ \ \ x_{4} =23\\\\\\Otvet; \ 2 \ ; \ 1 \ ; \ 23$

x = 0 - посторонний корень