Tính nhanh :
\(\frac{2005\times2004-1}{2003\times2005+2004}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
S
9
TN
16 tháng 7 2016
\(\frac{2005\cdot2004-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot\left(2003+1\right)-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2005-1}{2003\cdot2005+2004}\)
\(=\frac{2005\cdot2003+2004}{2003\cdot2005+2004}\)
\(=1\)
16 tháng 7 2016
2005 x 2004 - 1 / 2003 × 2005 + 2004
= 2005 × (2003 + 1) - 1 / 2003 × 2005 + 2004
= 2005 × 2003 + (2005 - 1) / 2003 × 2005 + 2004
= 2005 × 2003 + 2004 / 2003 × 2005 + 2004
= 1
DN
1
27 tháng 7 2015
\(\frac{2005\times2004-1}{2003\times2005+2004}=\frac{2005\times2003+2005-1}{2003\times2005+2004}=\frac{2005\times2003+2004}{2003\times2005+2004}=1\)
LT
1
XO
11 tháng 10 2020
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
11 tháng 10 2020
Ta có:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(P=\frac{1}{5}\cdot\left(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}\right)-\frac{2}{3}\cdot\left(\frac{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}{\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}}\right)\)
\(P=\frac{1}{5}-\frac{2}{3}=-\frac{7}{15}\)
LM
1
10 tháng 5 2015
Ta có: \(\frac{2004\cdot2007+6}{2005\cdot2005+2009}=\frac{\left(2005-1\right)\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-1\cdot2007+6}{2005\cdot2005+2009}=\frac{2005\cdot2007-2007+6}{2005\cdot2005+2009}\)
\(=\frac{\text{2005 x (2005 + 2) - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 2005 x 2 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\frac{\text{2005 x 2005 + 4010 - 2007 + 6}}{\text{2005 x 2005 + 2009}}=\text{ }\frac{\text{2005 x 2005 + 2009}}{\text{2005 x 2005 + 2009}}=1\)
23 tháng 12 2016
Bài 1:
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
\(\Rightarrow P=\frac{1\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2002}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
\(\Rightarrow P=\frac{1}{5}-\frac{2}{3}\)
\(\Rightarrow P=\frac{-7}{15}\)
Vậy \(P=\frac{-7}{15}\)
Bài 2:
Ta có: \(S=23+43+63+...+203\)
\(\Rightarrow S=13+10+20+23+...+103+100\)
\(\Rightarrow S=\left(13+23+...+103\right)+\left(10+20+...+100\right)\)
\(\Rightarrow S=3025+450\)
\(\Rightarrow S=3475\)
Vậy S = 3475
T
23 tháng 12 2016
1. \(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
=> P =\(\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{5\left(\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}\right)}-\frac{2\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}{3\left(\frac{1}{2002}+\frac{1}{2003}-\frac{1}{2004}\right)}\)
=> P = \(\frac{1}{5}-\frac{2}{3}\)
P = \(\frac{3}{15}-\frac{10}{15}\)
=> P =\(\frac{-7}{15}\)
2. ta có:
S = 23 + 43 + 63 +...+ 203
=> S = 13 + 10 + 23 + 20 +...+ 103 + 100
=> S = ( 13 + 23+...+ 103 ) + ( 10 + 20 +...+ 100 )
=> S = 3025 + 550
=> S = 3575
Vậy S = 3575
NT
1
12 tháng 2 2016
y=\(\frac{2006x2005-1}{2004x2006+2005}=\frac{2006x2005-1}{\left(2005-1\right)x2006+2005}=\frac{2006x2005-1}{2005x2006-2006+2005}=\frac{2006x2005-1}{2005x2006-1}=1\)
\(\frac{2005x2004-1}{2003x2005+2004}\)=\(\frac{4018019}{4018019}\)= 1
Bài giải
\(\frac{2005\text{ x }2004-1}{2003\text{ x }2005+2004}=\frac{2005\text{ x }2004-1}{2003\text{ x }2005+2005-1}=\frac{2005\text{ x }2004-1}{2005\text{ x }2004-1}=1\)