Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
B= 1/2 x 2/3 x 3/4 x ...........x 2002/2003 x 2003/2004
1 x 2 x 3 x 4 x .............x 2002 x 2003
2 x 3 x 4 x .............x 2003 x 2004
1
2004
a) 5 : 3/4 - 4 4/5 : 3/4
= 5 . 4/3 - 24/5 . 4/3
= (5 - 24/5) . 4/3
= 1/5 × 4/3
= 4/15
b) -3/5 . 2/7 + (-3/7) . 3/5 + (-3/7)
= (-3/7) . (2/5 + 3/5 + 1)
= (-3/7) . 2
= -6/7
c) [(-4 2/7) . 7/11 + 7/11 . (5 1/3)] . 5 - 5 2/3
= (-30/7 . 7/11 + 7/11 . 16/3) . 5 - 17/3
= (-30/11 + 112/33) . 5 - 17/3
= 2/3 . 5 - 17/3
= 10/3 - 17/3
= -7/3
d) 5/39 . [(7 4/5) . (1 2/3) + (8 1/3) . (7 4/5)]
= 5/39 . (39/5 . 5/3 + 25/3 . 39/5)
= 5/39 . 39/5 . (5/3 + 25/3)
= 1 . 10
= 10
=
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
a: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)
c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)
d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
a) Ta có: 1 + 5 + 6 = 12 ; 2 + 3 + 7 = 12
Vậy 1 + 5 + 6 = 2 + 3 + 7
b) Ta có:\(1^2\)\(+5^2\)\(+6^2\)\(=1+25+36=62\)
\(2^2\)\(+4^2\)\(+7^2\)\(=4+16+49=62\)
\(=>1^2\)\(+5^2\)\(+6^2\)\(=2^2\)\(+3^2\)\(+7^2\)
c) Ta có 1 + 6 +8 = 15; 2 + 4 + 9 = 15
Vậy 1 + 6 + 8 = 2 + 4 + 9
\(d,1^2\)\(+6^2\)\(+8^2\)\(=1+36+64=101\)
\(2^2\)\(+4^2\)\(+9^2\)\(=4+16+81=101\)
\(1^2\)\(+6^2\)\(+8^2\)\(=2^2\)\(+4^2\)\(+9^2\)
\(a,11\dfrac{3}{4}-\left(6\dfrac{5}{6}-4\dfrac{1}{2}\right)+1\dfrac{2}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{9}{2}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{41}{6}-\dfrac{27}{6}\right)+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{14}{6}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\dfrac{7}{3}+\dfrac{5}{3}\)
\(=\dfrac{47}{4}-\left(\dfrac{7}{3}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}-\dfrac{12}{3}\)
\(=\dfrac{47}{4}-4\)
\(=\dfrac{47}{4}-\dfrac{16}{4}\)
\(=\dfrac{31}{4}\)
c) Ta có: \(4\dfrac{3}{7}:\left(\dfrac{7}{5}\cdot4\dfrac{3}{7}\right)\)
\(=\dfrac{31}{7}:\left(\dfrac{7}{5}\cdot\dfrac{31}{7}\right)\)
\(=\dfrac{31}{7}:\dfrac{31}{5}\)
\(=\dfrac{5}{7}\)