Question

выполнить 3 задания начиная с 15го включительно с решением

Answer 1

Объяснение:

15.

$\displaystyle \\C_x^3+C_x^2=15*(x-1)$

ОДЗ: х≥3.

$\displaystyle \\\frac{x!}{(x-3)!*3!} +\frac{x!}{(x-2)!*2!}=15*(x-1)\\\\\\\frac{(x-3)!*(x-2)*(x-1)*x}{(x-3)!*1*2*3}+\frac{(x-2)!*(x-1)*x}{(x-2)!*1*2}=15*(x-1)\\\\\\\frac{(x-2)*(x-1)*x}{6} +\frac{(x-1)*x}{2} =15*(x-1)\ |:(x-1)\ \ (x\neq 1)\\\\\\\frac{(x-2)*x}{6} +(x-1)*x=15\ |*6\\\\x^2-2x+3x=90\\\\x^2+x-90=0\\\\x^2+10x-9x-90=0\\\\x*(x+10)-9*(x+10)=0\\\\(x+10)*(x-9)=0\\\\x+10=0\\\\x_1=-10\ \notin.\\\\x-9=0\\\\x_2=9.$

Ответ: х=9.

16.

$\displaystyle \\C_x^4=\frac{15*A_x^2}{4} \ |*4$

ОДЗ: х≥4.

$\displaystyle \\4*\frac{x!}{(x-4)!*4!}=15*\frac{x!}{(x-2)!} \\\\\\4*\frac{(x-4)!*(x-3)*(x-2)*(x-1)*x}{(x-4)!*1*2*3*4}=15*\frac{(x-2)!*(x-1)*x}{(x-2)!} \\\\\\\frac{(x-3)*(x-2)*(x-1)*x}{6} =15*(x-1)*x\ |:(x-1)*x\ \ \ (x\neq 1,\ \ x\neq 0)\\\\\\\frac{(x-3)*(x-2)}{6} =15\ |*6\\\\(x-3)*(x-2)=90\\\\x^2-3x-2x+6=90\\\\x^2-5x-84=0\\\\x^2-12x+7x-84=0\\\\x*(x-12)+7*(x-12)=0\\\\(x-12)*(x+7)=0\\\\x-12=0\\\\x_1=12.\\\\x+7=0\\\\x=-7\ \notin.$

Ответ: х=12.

17.

$\displaystyle \\C_{x-3}^2=21$

ОДЗ: х-3≥2     х≥5.

$\displaystyle \\\frac{(x-3)!}{(x-3-2)!*2!} =21\\\\\\\frac{(x-3)!}{(x-5)!*1*2}=21\\\\\\\frac{(x-5)!*(x-4)*(x-3)}{(x-5)!*2}=21\ |*2\\\\(x-4)*(x-3)=42\\\\x^2-4x-3x+12-42=0\\\\x^2-7x-30=0\\\\x^2-10x+3x-30=0\\\\x*(x-10)+3*(x-10)=0\\\\(x-10)*(x+3)=0\\\\x-10=0\\\\x_1=10.\\\\x+3=0\\\\x_2=-3\ \notin.$

Ответ: х=10.