Cho mình hỏi bài này làm như thế nào ạ?
View attachment 112901
Ta có:
[tex]\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}=\sqrt{\frac{a^2(a+1)^2+(a+1)^2+a^2}{a^2(a+1)^2}}=\sqrt{\frac{a^4+2a^2(a+1)+(a+1)^2}{a^2(a+1)^2}}\\ =\sqrt{\frac{(a^2+a+1)^2}{a^2(a+1)^2}}=\frac{a^2+a+1}{a(a+1)}=1+\frac{1}{a(a+1)}=1+\frac{1}{a}-\frac{1}{a+1}[/tex]
[tex]=>Q=(1+1-\frac{1}{2})+(1+\frac{1}{2}-\frac{1}{3})+...+(1+\frac{1}{n}-\frac{1}{n+1})+\frac{101}{n+1}\\ Q=n+(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1})+\frac{101}{n+1}\\ Q=n+1-\frac{1}{n+1}+\frac{101}{n+1}=n+1+\frac{100}{n+1}\geq 2\sqrt{(n+1)\frac{100}{(n+1)}}=20[/tex]
Dấu "=" xảy ra <=> n = 9
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