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@Ace Legona @ngonhuminh @Nguyễn Huy Tú @Đỗ Hương Giang @Tojimomi Katori Giúp mình với các bạn ơi 
áp dụng công thức này vào làm:
\(\dfrac{n}{a\left(a-n\right)}=\dfrac{1}{a-n}-\dfrac{1}{a}\)
Ta có: \(A=124\cdot\frac{1}{1984}\cdot\left(1-\frac{1}{1985}+\frac{1}{2}-\frac{1}{1986}+\frac{1}{3}-\frac{1}{1987}+...+\frac{1}{16}-\frac{1}{2000}\right)\)
\(\Rightarrow A=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\right)-\left(\frac{1}{1985}+\frac{1}{1986}+\frac{1}{1987}+...+\frac{1}{2000}\right)\right]\)
Laji cos: \(B=\frac{1}{16}\cdot\left(1-\frac{1}{17}+\frac{1}{2}-\frac{1}{18}+\frac{1}{3}-\frac{1}{19}+...+\frac{1}{1984}-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}-\frac{1}{17}-\frac{1}{18}-\frac{1}{19}-...-\frac{1}{2000}\right)\)
\(\Rightarrow B=\frac{1}{16}\cdot\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1984}\right)-\left(\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+...+\frac{1}{2000}\right)\right]\)
Lời giải:
Ta có:
\(f(x)=x^2+x\Rightarrow \frac{1}{f(x)}=\frac{1}{x^2+x}=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\)
Do đó:
\(\frac{1}{f(1)}=1-\frac{1}{2}\)
\(\frac{1}{f(2)}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{f(3)}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{f(2014)}=\frac{1}{2014}-\frac{1}{2015}\)
\(\frac{1}{f(2015)}=\frac{1}{2015}-\frac{1}{2016}\)
Cộng theo vế:
\(\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(2014)}+\frac{1}{f(2015)}=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(f\left(1\right)=-\dfrac{3}{2}.1=-\dfrac{3}{2}\)
\(f\left(-1\right)=-\dfrac{3}{2}.\left(-1\right)=\dfrac{3}{2}\)
\(f\left(2\right)=-\dfrac{3}{2}.2=-3\)
\(f\left(-2\right)=-\dfrac{3}{2}.\left(-2\right)=3\)
\(f\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{-3}{4}\)
\(f\left(-\dfrac{1}{2}\right)=-\dfrac{3}{2}.\left(-\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(f\left(a\right)< f\left(-a\right)\)