\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{2010^2-2009^2}{2009^2.2010^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2}=1-\frac{1}{2010^2}\)

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Ta có:

(n+1)²-n²=2n+1=n+(n+1)

=> A=\(\frac{2+1}{2^21^2}+\frac{2+3}{2^23^2}+... +\frac{2009+2010}{2009^22010^2}=1-\frac{1}{2^2}+\frac{1}{2^2} -\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2009^2}-\frac{1}{2010^2} <1 \)

Ta có : 

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\)\(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)

\(=\)\(\frac{2^2}{1^2.2^2}-\frac{1^2}{1^2.2^2}+\frac{3^2}{2^2.3^2}-\frac{2^2}{2^2.3^2}+\frac{4^2}{3^2.4^2}-\frac{3^2}{3^2.4^2}+...+\frac{10^2}{9^2.10^2}-\frac{9^2}{9^2.10^2}\)

\(=\)\(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)

\(=\)\(1-\frac{1}{10^2}\)

\(=\)\(\frac{100-1}{100}\)

\(=\)\(\frac{99}{100}\)

Chúc bạn học tốt ~ 

=3/1.4+5/4.9+7/9.16+......+19/81.100

=(1/1-1/4)+(1/4-1/9)+........+(1/81-1/100)

=1-1/100

=99/100<1(đpcm)

Xét số bất kì a. Ta sẽ chứng mỉnh (a + 1)² - a² = 2a + 1.

Thật vậy, ta có (a + 1)² - a² = a(a + 1) + (a + 1) - a² = (a² + a) + (a + 1) = 2a + 1 (đpcm).

Áp dụng ta có:

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)

\(=\frac{1}{1^2}-\frac{1}{10^2}< 1\left(đpcm\right)\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)

 \(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{99^2}-\frac{1}{100^2}=\frac{9999}{10000}\)