Question

bài 1 Rút gọn biểu thức:A=\(\sqrt{1+\dfrac{1}{a^2}+\dfrac{1}{\left(1+a\right)^2}}\)với a>0

2) Tính giá trị của tổng:

a) B =\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}\)+ \(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}\)+\(\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}\)+....+\(\sqrt{1+\dfrac{1}{2011^2}+\dfrac{1}{2012^2}}\)

Answer 1

Bài 1
\(1+\frac{1}{a^2}+\frac{1}{(a+1)^2}=(1+\frac{1}{a})^2-\frac{2}{a}+\frac{1}{(a+1)^2}\)

\(=(\frac{a+1}{a})^2-2.\frac{a+1}{a}.\frac{1}{a+1}+(\frac{1}{a+1})^2=(\frac{a+1}{a}-\frac{1}{a+1})^2\)

\(=(1+\frac{1}{a}-\frac{1}{a+1})^2\)

$\Rightarrow A=|1+\frac{1}{a}-\frac{1}{a+1}|=1+\frac{1}{a}-\frac{1}{a+1}$ với $a>0$

 

Answer 2

Bài 2:

Áp dụng kết quả bài 1 thì:
\(B=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2011}-\frac{1}{2012}\)

\(=2011+(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011})-(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012})\)

\(=2012-\frac{1}{2012}\)