Question

17. Решите уравнение ​

Answer 1

Ответ: x=13.

Объяснение:

$\displaystyle\\\frac{C_x^3+C_x^4}{C_{x+1}^2}=11\\\\\\\frac{\frac{x!}{(x-3)!*3!}+\frac{x!}{(x-4)!*4!} }{\frac{(x+1)!}{(x+1-2)!*2!} } =11$

$\displaystyle\\\frac{\frac{(x-3)!*(x-2)*(x-1)*x}{(x-3)!*1*2*3} +\frac{(x-4)!*(x-3)*(x-2)*(x-1)*x}{(x-4)!*1*2*3*4} }{\frac{(x-1)!*x*(x+1)}{(x-1)!*1*2} } =11$

$\displaystyle\\\frac{\frac{(x-2)*(x-1)*x}{6}+\frac{(x-3)*(x-2)*(x-1)*x}{24} }{\frac{x*(x+1)}{2} } }=11$

$\displaystyle\\\frac{\frac{(x-2)*(x-1)}{6}+\frac{(x-3)*(x-2)*(x-1)}{24} }{\frac{x+1}{2} }=11$

$\displaystyle\\\frac{\frac{(x-2)*(x-1)}{3} +\frac{(x-3)*(x-2)*(x-1)}{12} }{x+1} =11$

$\displaystyle\\\frac{4*(x-2)*(x-1)+(x-3)*(x--2)*(x-1)}{12*(x+1)} =11$

$\displaystyle\\\frac{(x-2)*(x-1)*(4+(x-3))}{x+1} =11*12\\\\\frac{(x-2)*(x-1)*(x+1)}{x+1} =132\\\\(x-2)*(x-1)=132\\\\x^2-3x+2-132=0\\\\x^2-3x-130=0\\\\x^2-13x+10x-130=0\\\\x*(x-13)+10*(x-13)=0\\\\(x-13)*(x+10)=0\\\\x-13=0\\\\x_1=13\\\\x+10=0\\\\x_2=-10\notin.$