Question

Rozwiąż równanie:
a) 10*11!-13!=(1-n)(14*10!+12!)

b) [tex]\frac{(n+1)!-2n!}{(n+2)!} =\frac{1}{10}[/tex]

Answer 1

a)

[tex]10\cdot 11!-13!=(1-n)(14\cdot10!+12!)\\11!(10-12\cdot13)=(1-n)10!(14+11\cdot12)\\11(10-156)=(1-n)(14+132)\\11\cdot(-146)=(1-n)\cdot 146\\-11=1-n\\n=12[/tex]

b)

[tex]\dfrac{(n+1)!-2n!}{(n+2)!}=\dfrac{1}{10}\\\\\dfrac{(n+1)!}{(n+2)!}-\dfrac{2n!}{(n+2)!}=\dfrac{1}{10}\\\\\dfrac{1}{n+2}-\dfrac{2}{(n+1)(n+2)}=\dfrac{1}{10}\Big|\cdot 10(n+1)(n+2)\\\\10(n+1)-2\cdot10=(n+1)(n+2)\\10n+10-20=n^2+2n+n+2\\n^2-7n+12=0\\n^2-3n-4n+12=0\\n(n-3)-4(n-3)=0\\(n-4)(n-3)=0\\n=4 \vee n=3[/tex]