Đặt 2 là thừa số chung rồi tính tổng 1+2+3+4+...+25 ( theo công thức )

\(=2\left(1+2+3+4+...+25\right)\)
\(=2.\dfrac{\left(1+25\right).25}{2}\)
\(=\left(1+25\right).25\)
\(=26.25\)
\(=650\)

cách của mình từng đc olm lụa chọn đó

Có A= 2.1+ 2.3+2.5+........+2. 97+ 2.99 

=>A=2 . (1+3+5+.....+97+99)

=>A=2 . [(1+99)+(3+97)+(5+95)+.....+(50+50)]

=>A=2 . [100+100+100+....+100]   Có 25 số 100

=>A=2 . [100.25]

=>A= 2 . 2500

=>A=5000

Câu b bạn làm tương tự nha

Bạn tham khảo lời giải tại đây:

https://olm.vn/hoi-dap/detail/81621153379.html

\(\left(9!-8!\right).7!.x=1^2.2^2.3^2.4^2.....8^2\)

\(\Leftrightarrow\)\(8!\left(9-1\right).7!.x=\left(1.2.3.4.....8\right).\left(1.2.3.4.....8\right)\)

\(\Leftrightarrow\)\(8!.8.7!.x=8!.8!\)

\(\Leftrightarrow\)\(8!.8!.x=8!.8!\)

\(\Leftrightarrow\)\(x=\frac{8!.8!}{8!.8!}\)

\(\Leftrightarrow\)\(x=1\)

Vậy \(x=1\)

\(\left(9!-8!\right).7!.x=1^2.2^2.3^2.4^2.....8^2\)

\(\Leftrightarrow\)\(8!\left(9-1\right).7!.x=\left(1.2.3.4.....16\right).\left(1.2.3.4.....16\right)\)

\(\Leftrightarrow\)\(8!.8.7!.x=8!.8!\)

\(\Leftrightarrow\)\(8!.8!.x=8!.8!\)

\(\Leftrightarrow\)\(x=\frac{8!.8!}{8!.8!}\)

\(\Leftrightarrow\)\(x=1\)

Vậy \(x=1\)

Vậy 

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

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

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

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

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

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

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

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

Cho \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}\)... là A, ta có:

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

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

A = 1 \(-\frac{1}{10^2}\) <1

Vậy: A < 1

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

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

=1-1/100<1

Vậy tổng trên <1