A<1

bạn tính phần mẫu ra rồi làm như dạng sai phân bình thường

i nhanh và đúng mk k cho nhé, mk hứa

a bài 1 hay 2

bạn phải làm chứ k làm mk k có k

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A=3 /1^2.2^2 +5 / 2^2.3^2 +7/3^2.4^2 +...+ 19 /9^2.10^2

=1/1^2-1/2^2+1/2^2-1/3^2+1/3^2-1/4^2+....+1/9^2-1/10^2

=1/1^2-1/10^2

=99/100

=0,99

vậy A< 1

ta có:

 \(\frac{1}{11}\)>\(\frac{10}{20}\)

\(\frac{1}{12}\)>\(\frac{10}{20}\)

\(\frac{1}{13}\)>\(\frac{10}{20}\)

....

\(\frac{1}{19}\)>\(\frac{10}{20}\)

=>E >\(\frac{10}{20}\)

vậy E > \(\frac{1}{2}\)

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:

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

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

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

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

Ta có:

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

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

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

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

Vậy

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

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

\(=\frac{3}{1.4}+\frac{5}{4.9}+.....+\frac{19}{81.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{100}\)

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

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

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

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

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

Ta có:

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

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

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

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

\(\RightarrowĐPCM\)

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