ĐẶT A=\(\frac{1}{3^0}+\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

  \(\frac{1}{3}A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)

\(\frac{1}{3}A-A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)

\(\frac{-2}{3}A=\frac{1}{3^{2006}}-\frac{1}{3^0}\)

\(A=\frac{\frac{1}{3^{2006}}-1}{\frac{-2}{3}}\)

\(3S=3+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(3S-S=\left(3+\frac{1}{3}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\right)\)

\(2S=3-\frac{1}{3^{2005}}\)

\(2S=\frac{3^{2006-1}}{3^{2005}}\)

\(S=\frac{3^{2006}-1}{3^{2005}.2}\)

S = 1/3 + 1/3² + 1/3³ + ... + 1/3²⁰⁰⁵

=> 3S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴

=> 3S - S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴ - (1/3 + 1/3² + 1/3³ + ... + 1/3²⁰⁰⁵)

=> 2S = 1 + 1/3 + 1/3² + ... + 1/3²⁰⁰⁴ - 1/3 - 1/3² - 1/3³ - ... - 1/3²⁰⁰⁵

=> 2S = 1 - 1/3²⁰⁰⁵

=> S = \(\frac{\frac{1}{3^{2005}}}{2}\)

=> S = 1/3²⁰⁰⁵.2

\(P=1+5+5^2+............+5^{2005}\)

\(5P=5+5^2+5^3+...........5^{2006}\)

\(5P-P=5^{2006}-1\)

\(P=\frac{5^{2006}-1}{4}\)

\(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)

Ta nhận thấy các cặp số đều bằng 3/5 và các dấu cũng giống nhau. ( các số có cùng dấu thì phân số đó cũng cùng dấu.)

=> Phân số này sẽ bằng 3/5

\(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\)

Ta nhận thấy các cặp số đều bằng -3/5 và các dấu thì trái nhau.  ( các số có trái dấu thì phân số đó cũng trái dấu.)

=> Phân số này sẽ bằng -3/5.

Sau khi rút gọn bài toán sẽ thành:

\(\left(\frac{3}{5}-\frac{3}{5}\right)\div\frac{1890}{2005}+115=115\)

Câu b tạm thời mình chưa nghĩ ra. Chúc bạn học tốt.

a)  \(A=\left(\frac{3}{5}-\frac{3}{5}\right):\frac{1890}{2005}+115\)

\(\Rightarrow A=115\)

b)   \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)

\(\Rightarrow3B-B=\left(1+\frac{1}{3}+....+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2B=1-\frac{1}{3^{2005}}\)

\(\Rightarrow B=\frac{1-\frac{1}{3^{2005}}}{2}\)

\(\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)

 \(\Rightarrow B< \frac{1}{2}\)

Đặt A \(=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\)

\(\Rightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2004}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{2005}}\)

\(\Rightarrow A=\left(1-\frac{1}{3^{2005}}\right):2\)

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(\Leftrightarrow3A-A=1-\frac{1}{3^{2005}}\)

\(\Leftrightarrow A=\frac{1-\frac{1}{3^{2005}}}{2}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)

\(3A=1+\frac{1}{3}+...+\frac{1}{3^{2004}}\)

\(3A-A=1-\frac{1}{3^{2005}}\)

\(A=\frac{1-\frac{1}{3^{2005}}}{2}\)

\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}+2\left(1.\frac{1}{n}-1.\frac{1}{n+1}-\frac{1}{n}.\frac{1}{n+1}\right)=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\); vì \(\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n\left(n+1\right)}=0\)

\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{2005}-\frac{1}{2006}\right)\)

\(=2005+1-\frac{1}{2006}=2005\frac{2005}{2006}\)

a) \(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}\)

\(\Rightarrow3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2014}}\)

\(\Rightarrow3B-B=1-\frac{1}{3^{2015}}\)

\(B=\frac{1-\frac{1}{3^{2015}}}{2}\)

giúp câu P luôn với bạn

b) Đặt \(A=\frac{2004}{1}+\frac{2003}{2}+\frac{2002}{3}+...+\frac{1}{2004}\)

\(\Rightarrow A=\left(\frac{2003}{2}+1\right)+\left(\frac{2002}{3}+1\right)+...+\left(\frac{1}{2004}+1\right)+1\) ( tách 2004/1=2004 ra, cộng cho các phân số kia mỗi phân số 1 đơn vị, thì còn dư ra 1)

\(A=\frac{2005}{2}+\frac{2005}{3}+...+\frac{2005}{2004}+\frac{2005}{2005}\) ( 1 = 2005/2005)

\(A=2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)\)

Thay A vào P được

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2005}}{2005.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}+\frac{1}{2005}\right)}\)

\(P=\frac{1}{2005}\)