3a.
Đề bài câu này sai, đề đúng phải là \(M>1\)
Do: \(\dfrac{a}{b+c}>\dfrac{a}{a+b+c}\) ; \(\dfrac{b}{a+c}>\dfrac{b}{a+b+c}\) ; \(\dfrac{c}{a+b}>\dfrac{c}{a+b+c}\)
\(\Rightarrow M>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\)
b.
Ta có: \(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)
Áp dụng:
\(S=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{99^2}\right)\)
\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{98.100}{99^2}\)
\(=\dfrac{1.2.3...98}{2.3.4...99}.\dfrac{3.4.5...100}{2.3.4...99}\)
\(=\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{50}{99}\)
