tính nhanh
\(\frac{2003x1999-2003x999}{2004x999x1994}\)
K
Khách
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B
1
23 tháng 7 2018
\(\frac{2003\cdot1999-2003\cdot999}{2004\cdot999+1004}\)
\(=\frac{2003\cdot\left(1999-999\right)}{2004\cdot\left(999+1\right)}\)
\(=\frac{2003\cdot1000}{2004\cdot1000}\)
\(=\frac{2003}{2004}\)
NT
1
30 tháng 12 2022
\(\dfrac{2003\times1999-2003\times999}{2004\times999+1004}=\) \(\dfrac{2003\times\left(1999-999\right)}{\left(1000+1004\right)\times999+1004}\)
\(=\dfrac{2003\times1000}{1000\times999+1004\times999+1004}\)
\(=\dfrac{2003\times1000}{1000\times999+1004\times\left(999+1\right)}\)
\(=\dfrac{2003\times1000}{1000\times999+1004\times1000}\)
\(=\dfrac{2003\times1000}{1000\times\left(999+1004\right)}\)
\(=\dfrac{2003}{2003}=1\)
24 tháng 8 2016
2003x1999-2003x999
= 2003x(1999 - 999 )
= 2003 x 1000
= 2003000
2004 x 999 + 1004
= ( 2003 + 1 ) x 999 + 1004
= 2003 x 999 + 999 + 1004
= 2003 x 999 +2003
= 2003 x ( 999 + 1 )
= 2003 x 1000
= 2003000
24 tháng 8 2016
\(\frac{2003\times1999-2003\times999}{2004\times999+1004}\)
\(=\frac{2003\times\left(1999+999\right)}{2004\times999+1004}\)
\(=\frac{2003\times2998}{2001996+1004}\)
\(=\frac{60004994}{2003000}\)
HT
2
25 tháng 4 2018
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
25 tháng 4 2018
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
CC
2
29 tháng 7 2016
\(\frac{1\times2\times25}{6\times5\times7}\)\(=\frac{1\times2\times5\times5}{2\times3\times5\times7}=\frac{5}{3\times7}=\frac{5}{21}\)
LP
29 tháng 7 2016
1/6×2/5×25/7=
rut gon:25(tử)÷5 (mâu), 6 (tử)÷2 (mau)
được
1/3×1/1×5/7=5/21
16 tháng 5 2019
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
16 tháng 5 2019
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)-\frac{1}{9}\)
\(=1-\frac{1}{9}=\frac{8}{9}\)
~ Hok tốt ~
\(\frac{2003x1999-2003x999}{2004x999x1994}=\frac{2003x\left(1999-999\right)}{2004x999x1994}\)
\(=\frac{2003x1000}{2004x999x1994}=\frac{1}{1994x}\)
\(\Rightarrow❤️✔️✨♕✨✔️❤️\Leftarrow\)
\(\text{Bài làm :}\)
\(\frac{2003\cdot1999-2003\cdot999}{2004\cdot999\cdot1994}=\frac{2003\cdot1999}{999\cdot1994}=\frac{4003997}{1992006}\)
\(\text{Chúc bạn học tốt !}\)