Question

СРОЧНО!!!
выполнить задания 18,19,20 с решением

Answer 1

Объяснение:

18.

$\displaystyle \\C_{x+1}^5=\frac{3A_x^3}{8}$

ОДЗ: х+1≥5     х≥4.

$\displaystyle \\\frac{(x+1)!}{(x+1-5)!*5!}=\frac{3}{8} *\frac{x!}{(x-3)!} \\\\\\\frac{(x+1)!}{(x-4)!*1*2*3*4*5}=\frac{3}{8}*\frac{(x-3)!*(x-2)*(x-1)*x}{(x-3)!} \\\\\\\frac{(x-4)!*(x-3)*(x-2)*(x-1)*x*(x+1)}{(x-4)!*120} =\frac{3}{8}*(x-2)*(x-1)*x\ |*8\\\\\\\frac{(x-3)*(x-2)*(x-1)*x*(x+1)}{15} =3*(x-2)*(x-1)*x\ |:(x-2)*(x-1)*x\\\\\\\frac{(x-3)*(x+1)}{15} =3\ |*15\\\\x^2-3x+x-3=45\\\\x^2-2x-48=0$

$x^2-8x+6x-48=0\\\\x*(x-8)+6*(x-8)=0\\\\(x-8)*(x+6)=0\\\\x-8=0\\\\x_1=8.\\\\x+6=0\\\\x_2=-6\notin.$

Ответ: х=8.

19.

$\displaystyle \\\left \{ {{A_x^y=10*A_x^{y-1}\ \ \ \ (1)} \atop {C_x^y=\frac{5}{3} *C_x^{y-1}\ \ \ \ (2)}} \right.$

Упростим уравнение (1):

$\displaystyle \\\frac{x!}{(x-y)!} =10*\frac{x!}{(x-(y-1))!} \\\\\\\frac{x!}{(x-y)!}=10*\frac{x!}{(x-y+1)!} \ |:x! \\\\\\\frac{1}{(x-y)!}=\frac{10}{(x-y)!*(x-y+1)}\ |:(x-y)!\\\\\\1=\frac{10}{x-y+1}\\\\x-y+1=10\\\\x-y=9\\\\y=x-9.$

Упростим уравнение (2):

$\displaystyle \\\frac{x!}{(x-y)!*y!}=\frac{5}{3}*\frac{x!}{(x-(y-1))!*(y-1)!} \ | :x! \\\\\\\frac{1}{(x-y)!*(y-1)!*y}=\frac{5}{3} * \frac{1}{(x-y+1)!*(y-1)!}\ | *(x-y)!*(y-1)!\\\\\\\frac{1}{y} =\frac{5}{3} *\frac{(x-y)!}{(x-y)!*(x-y+1)} \ |*3y\\\\\\3=\frac{5*y}{x-y+1}\\\\3*(x-y+1)=5y\\\\3x-3y+3=5y\\\\3x-8y=-3\ \ \ \ \ \ \Rightarrow\\\\3x-8*(x-9)=-3\\\\3x-8x+72=-3\\\\-5x=-75\ |:(-5)\\\\x=15.\ \ \ \ \ \ \Rightarrow\\\\y=15-9\\\\y=6.$

Ответ: х=15,   у=6.