Question

Пожалуйста быстрее решите

Answer 1

$\displaystyle\bf\\1)\\\\\frac{x}{A_{x}^{3} }=\frac{1}{2} }\\\\\\A_{x} ^{3} =\frac{x!}{(x-3)!} =\frac{(x-3)!\cdot(x-2)(x-1)x}{(x-3)!} =(x-2)(x-1)x\\\\\\\frac{x}{(x-2)(x-1)x} =\frac{1}{2} \\\\\\\frac{1}{(x-2)(x-1)}=\frac{1}{2} \\\\\\(x-2)(x-1)=2\\\\\\x^{2} -x-2x+2-2=0\\\\\\x^{2} -3x=0\\\\\\x(x-3)=0\\\\\\\left[\begin{array}{ccc}x_{1} =0-neyd\\x_{2} =3\end{array}\right\\\\\\Otvet:3$

$\displaystyle\bf\\2)\\\\\frac{A_{m+n} ^{n+2} +A_{m+n}^{n+1} }{A_{m+n}^{n} } \\\\\\A_{m+n} ^{n+2} =\frac{(m+n)!}{(m+n-n-2)!} =\frac{(m+n)!}{(m-2)!} \\\\\\A_{m+n} ^{n+1} =\frac{(m+n)!}{(m+n-n-1)!} =\frac{(m+n)!}{(m-1)!} \\\\\\A_{m+n} ^{n} =\frac{(m+n)!}{(m+n-n)!} =\frac{(m+n)!}{m!} \\\\\\\frac{(m+n)!}{(m-2)!} +\frac{(m+n)!}{(m-1)!} =\frac{(m+n)!}{(m-2)!} +\frac{(m+n)!}{(m-2)!\cdot(m-1)} =\\\\\\=\frac{(m+n)!\cdot(m-1+1)}{(m-2)!\cdot(m-1)} =\frac{(m+n)!\cdot m}{(m-2)!\cdot(m-1)}$

$\displaystyle\bf\\\frac{(m+n)!\cdot m}{(m-2)!\cdot(m-1)}:\frac{(m+n)!}{m!}= \frac{(m+n)!\cdot m}{(m-2)!\cdot(m-1)}\cdot\frac{m!}{(m+n)!} =\\\\\\=\frac{m\cdot(m-2)!\cdot(m-1)\cdot m}{(m-2)!\cdot(m-1)} =m^{2}$