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\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\)\(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(a+b+c\right)}{abc}}\)=\(\sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}=\)\(|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}=\sqrt{1+\frac{1}{2^2}+\frac{1}{\left(-3\right)^2}}\)\(=|\frac{1}{1}+\frac{1}{2}+\frac{1}{-3}|=1+\frac{1}{2}-\frac{1}{3}\)
Tương tự ta có M=\(1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{99}-\frac{1}{100}\)=\(98+\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)
Xét \(P=\sqrt{\frac{1}{1^2}+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\) với a>0
\(P^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}\)
\(=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\left(\frac{a^2+a+1}{a\left(a+1\right)}\right)^2\)
Do a>o nên \(P=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
Áp dụng kết quả của P ta có:
\(A=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}+\frac{1}{3}\right)+....+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\) \(A=2012+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{2012}-\frac{1}{2013}\right)\)
\(A=2012+1-\frac{1}{2013}\)
\(A=2013-\frac{1}{2013}=\frac{4052168}{2013}\)
Vậy \(A=\frac{4052168}{2013}\)
\(C=\frac{T}{M}\)
\(M=\left(1+\frac{3998}{2}\right)+\left(1+\frac{3997}{3}\right)+.....+\left(1+\frac{1}{3999}\right)+\frac{4000}{4000}\)
\(=\frac{4000}{2}+\frac{4000}{3}+......+\frac{4000}{3999}+\frac{4000}{4000}=4000.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4000}\right)\)
\(=4000.T\)
\(C=\frac{T}{M}=\frac{T}{4000T}=\frac{1}{4000}\)