Tính nhanh:
E = 2/3.5 + 7/5.12 + 9/4.39
F = 1/2003.2002 - 1/2002.2001 - 1/2001.2000 - ..... - 1/3.2 - 1/2.1
H = 1/13 + 3/13.23 + 3/23.33 + ..... + 3/1993.2003
Bài 1 Tính nhanh
A:75-6/13+3/17-3/19|275--22/13+11/17-11/19
B:1/13+3/13.23+3/23.33+3/33.43+...+3/1993.2003
C=-1/2003.2002-1/2002.2001-1/2001.2000-...-1/3.2-1/2.1
D=2/3.5+7/5.12+9/4.39
Bài 1 Tính nhanh
A:75-6/13+3/17-3/19|275--22/13+11/17-11/19
B:1/13+3/13.23+3/23.33+3/33.43+...+3/1993.2003
C=-1/2003.2002-1/2002.2001-1/2001.2000-...-1/3.2-1/2.1
D=2/3.5+7/5.12+9/4.39
Bài 1 Tính nhanh
A:75-6/13+3/17-3/19|275--22/13+11/17-11/19
B:1/13+3/13.23+3/23.33+3/33.43+...+3/1993.2003
C=-1/2003.2002-1/2002.2001-1/2001.2000-...-1/3.2-1/2.1
-1/2003.2002-1/2002.2001-1/2001.2000-...-1/3.2-1/2.1
\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\)
\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-1\left(1-\frac{1}{2003}\right)\)
\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)
\(=-1\cdot\frac{2002}{2003}\)
\(=-\frac{2002}{2003}\)
Tính: A=\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{2003}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2001}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{2003}-1\right)=-\left(-\frac{2002}{2003}\right)=\frac{2002}{2003}\)
Vậy ....
\(\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+.......................+\frac{3}{1993.2003}\)
\(=\frac{3}{3.13}+\frac{3}{13.23}+...+\frac{3}{1993.2003}\)
\(=\frac{1}{10}.\left(1-\frac{3}{13}+\frac{3}{13}-\frac{3}{23}+...+\frac{3}{1993}-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\left(1-\frac{3}{2003}\right)\)
\(=\frac{1}{10}.\frac{2000}{2003}\)
\(=\frac{200}{2003}\)
Đặt \(A=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(\Rightarrow A=3\left(\frac{1}{3.13}+\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)\)
\(\Rightarrow A=\frac{3}{10}.\left(\frac{2003}{6009}-\frac{3}{6009}\right)\)
\(\Rightarrow A=\frac{3}{10}.\frac{2000}{6009}\)
\(\Rightarrow A=\frac{200}{2003}\)
Bài 3 Thực hiện phép tính một cách hợp lí
N=\(\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(N=\frac{1}{13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(=\frac{3}{3.13}+\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
\(=\frac{3}{10}\left(\frac{10}{3.13}+\frac{10}{13.23}+\frac{10}{23.33}+..+\frac{10}{1993.2003}\right)\)
\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\)
\(=\frac{3}{10}\left(\frac{1}{3}-\frac{1}{2003}\right)=\frac{3}{10}.\frac{2000}{6009}=\frac{200}{2003}\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{3}{13.23}\)\(+\)\(\frac{3}{23.33}\)\(+...+\)\(\frac{3}{1993.2003}\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13.23}+\frac{1}{23.33}+...+\frac{1}{1993.2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}\left(\frac{1}{13}-\frac{1}{2003}\right)\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\left[\frac{3}{10}.\frac{1990}{26039}\right]\)
\(N=\)\(\frac{1}{13}\)\(+\)\(\frac{597}{26039}\)
\(N=\)\(\frac{200}{2003}\)
dễ vãi lồn làm làm đéo j phá đê
Tính
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-...-\frac{1}{2.3}-\frac{1}{2.1}\)
giúp mình với nha
Bạn tham khảo ở lcik này ! Mình mới trả lời ở đó !
Câu hỏi của Nguyễn Thị Ngọc Ánh - Toán lớp 7 - Học toán với OnlineMath
https://olm.vn/hoi-dap/detail/228829251573.html
tính nhanh:
\(\frac{1}{13}\)+\(\frac{3}{13.23}\)+\(\frac{3}{23.33}\)+...+\(\frac{3}{1993.2003}\)
Giúp mik nhoa, mik đang cần gấp lắm ^-^
\(\frac{1}{13}+\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\)
\(=\frac{1}{13}+\left[\frac{3}{13\cdot23}+\frac{3}{23\cdot33}+...+\frac{3}{1993\cdot2003}\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13\cdot23}+\frac{1}{23\cdot33}+...+\frac{1}{1993\cdot2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\left[\frac{1}{13}-\frac{1}{2003}\right]\right]\)
\(=\frac{1}{13}+\left[\frac{3}{10}\cdot\frac{1990}{26039}\right]\)
\(=\frac{1}{13}+\frac{597}{26039}\)
\(=\frac{200}{2003}\)
Đặt A= 1/13 + 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
A- 1/13 = 3/13.23 + 3/ 23.33 + ... + 3/1993.2003
10/3 ( A-1/3) = 10/3. (3/13.23 + 3/ 23.33 + ... + 3/1993.2003)
10/3A - 10/9 = 10/13.23 + 10/ 23.33 + ... + 10/1993.2003
10/3A - 10/9 = 1/13 - 1/23 + 1/23 - 1/33 +...+ 1/1993- 1/2003
10/3A = 1/13 - 1/2003 + 10/9
10/3 A= ?
đến đây bn tự làm nha
10/3A - 10/9 = 1/13
Ta loại \(\frac{1}{13}\)ra khỏi biểu thức để dễ tính hơn, sau đó cộng vào.
Gọi A là biểu thức \(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\)
Gấp A lên \(\frac{10}{3}\)lần, ta có :
\(A\times\frac{10}{3}=\frac{10}{3}\times\left(\frac{3}{13.23}+\frac{3}{23.33}+...+\frac{3}{1993.2003}\right)\)
\(A\times\frac{10}{3}=\frac{10}{13.23}+\frac{10}{23.33}+...+\frac{10}{1993.2003}\)
\(A\times\frac{10}{3}=\frac{1}{13}-\frac{1}{23}+\frac{1}{23}-\frac{1}{33}+...+\frac{1}{1993}-\frac{1}{2003}\)
\(A\times\frac{10}{3}=\frac{1}{13}-\frac{1}{2003}\)
\(A\times\frac{10}{3}=\frac{1990}{26039}\)
\(A=\frac{1990}{26039}\div\frac{10}{3}\)
\(A=\frac{597}{26039}\)
Biểu thức trên = \(\frac{597}{26039}+\frac{1}{13}=\frac{200}{2003}\)