IDNLearn.com is your go-to resource for finding precise and accurate answers. Get accurate and comprehensive answers from our network of experienced professionals.
Sagot :
[tex] \frac{(15-x)!}{(13-x)!\cdot2!}=\frac{x!}{(x-2)!\cdot2!}\\
\frac{(14-x)(15-x)}{2}=\frac{(x-1)x}{2}\\
(14-x)(15-x)=(x-1)x\\
210-14x-15x+x^2=x^2-x\\
28x=210\\
x=\frac{210}{28}=7.5[/tex]
The factorial is defined for natural numbers and since 7.5 isn't one, this equation has no solutions.
The factorial is defined for natural numbers and since 7.5 isn't one, this equation has no solutions.
[tex](15-x)!=[15-(x+2)]!\times[15-(x+1)]\times(15-x)\\\\=(15-x-2)!\times(15-x-1)\times(15-x)\\\\=(13-x)!\times(14-x)\times(15-x)\\\\=(13-x)!\times(210-14x-15x+x^2)\\\\=(13-x)!\times(x^2-29x+210)\\\\therefore:\frac{(15-x)!}{(13-x)!\cdot2!}=\frac{(13-x)!\cdot(x^2-29x+210)}{(13-x)!\cdot2!}=\frac{x^2-29x+210}{2}[/tex]
[tex]x!=(x-2)!\times(x-1)\times x=(x-2)!\times(x^2-x)\\\\therefore:\frac{x!}{(x-2)!\cdot2!}=\frac{(x-2)!\cdot(x^2-x)}{(x-2)!\cdot2}=\frac{x^2-x}{2}\\\\======================================\\\\\frac{x^2-29x+210}{2}=\frac{x^2-x}{2}\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\\\x^2-29x+210=x^2-x\\\\x^2-x^2-29x+x=-210\\\\-28x=-210\ \ \ \ \ |divide\ both\ sides\ by\ (-28)\\\\x=7.5\notin\mathbb{Z}[/tex]
[tex]This\ equation\ has\ not\ solution\ because\ x\ must\ be\ an\ integer!\\\\x\in\O[/tex]
[tex]x!=(x-2)!\times(x-1)\times x=(x-2)!\times(x^2-x)\\\\therefore:\frac{x!}{(x-2)!\cdot2!}=\frac{(x-2)!\cdot(x^2-x)}{(x-2)!\cdot2}=\frac{x^2-x}{2}\\\\======================================\\\\\frac{x^2-29x+210}{2}=\frac{x^2-x}{2}\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\\\x^2-29x+210=x^2-x\\\\x^2-x^2-29x+x=-210\\\\-28x=-210\ \ \ \ \ |divide\ both\ sides\ by\ (-28)\\\\x=7.5\notin\mathbb{Z}[/tex]
[tex]This\ equation\ has\ not\ solution\ because\ x\ must\ be\ an\ integer!\\\\x\in\O[/tex]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.
