Đặt \(A=\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{602.605}\)

\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)

\(A=\frac{1}{5}-\frac{1}{605}\)

\(A=\frac{24}{121}\)

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{602\cdot605}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\frac{1}{5}-\frac{1}{605}+0+...+0\)

\(=\frac{24}{121}\)

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\)\(\frac{3}{11\cdot14}+...+\)\(\frac{3}{602\cdot605}\)

\(=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{602\cdot605}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}\)\(-\frac{1}{605}\)

\(=\frac{1}{5}-\frac{1}{605}\)

\(=\frac{121}{605}-\frac{1}{605}\)

\(=\frac{120}{605}=\frac{24}{121}\)

Bài này dùng công thức nhé 

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{602.605}\)

\(=\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\)\(\frac{1}{5}-\frac{1}{605}\)

\(=\)\(\frac{24}{121}\)

Chúc bạn học tốt ~ 

= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 -  7 + 8 ) + ....+ ( 601 - 602 - 603 + 604 ) 

= 0  + 0 + .. + 0 

= 0 

=1-2-3+4)+(5-6-7+8)+...+(601-602-603+6004)

=0+0+...+0

=0 

k mik nha!

1 - 2 - 3 + 4 + 5- 6 - 7 + .... + 601 - 602 - 603 + 604 

= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (601 -602 - 603 + 604)

= 0 + 0 + ... + 0 

=  0 

a)

Trung bình cộng n-1 số còn lại là:

              (1+2+3+…+(n-2)+(n-1)):(n-1)=\(\frac{602}{7}\)

Mà 1+2+3+…+(n-2)+(n-1)=\(\frac{\left(n-1\right).\left(n-1+1\right)}{2}=\frac{\left(n-1\right).n}{2}\)

=>\(\left(1+2+3+...+\left(n-2\right)+\left(n-1\right)\right):\left(n-1\right)=\frac{\left(n-1\right).n}{2}:\left(n-1\right)=\frac{602}{7}\)

=>\(\frac{\left(n-1\right).n:\left(n-1\right)}{2}=\frac{602}{7}\)

=>\(\frac{n}{2}=\frac{602}{7}=>\frac{7n}{14}=\frac{1204}{14}=>7n=1204=>n=172\)

Tìm số bị xóa chứ không phải tìm số n.