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246 x 2005 - 2005 x 148 + 2005 x 102
= 2005 x (246 - 148 + 102)
= 2005 x 200
= 401000
169 : 3 + 562 : 3 + 169 : 3
= (169 + 562 + 169) : 3
= 900 : 3
= 300
201 x 2 + 201 + 201 x 3 + 201 x 4
= 201 x 2 + 201 x 1 + 201 x 3 + 201 x 4
= 201 x (2 + 1 + 3 + 4)
= 201 x 10
= 2010
\(\frac{4}{7}\times\frac{5}{6}+\frac{4}{7}\times\frac{1}{6}\)
\(=\frac{4}{7}\times\left(\frac{5}{6}+\frac{1}{6}\right)\)
\(=\frac{4}{7}\times1\)
\(=\frac{4}{7}\)
\(\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)
\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)
\(=\frac{7}{9}\times1\)
\(=\frac{7}{9}\)
\(\frac{3}{5}\times\frac{7}{9}-\frac{3}{5}\times\frac{2}{9}\)
\(=\frac{3}{5}\times\left(\frac{7}{9}-\frac{2}{9}\right)\)
\(=\frac{3}{5}\times\frac{5}{9}\)
\(=\frac{1}{3}\)
\(\frac{2}{5}\times\frac{3}{4}\times\frac{3}{5}=\frac{3}{10}\times\frac{3}{5}=\frac{9}{50}\)
\(\frac{5}{9}\times\frac{1}{4}+\frac{4}{9}\times\frac{3}{12}\)
\(=\frac{5}{9}\times\frac{1}{4}+\frac{4}{9}\times\frac{1}{4}\)
\(=\frac{1}{4}\times\left(\frac{5}{9}+\frac{4}{9}\right)\)
\(=\frac{1}{4}\times1\)
\(=\frac{1}{4}\)
\(\frac{2006}{2005}\times\frac{3}{4}-\frac{3}{4}\times\frac{1}{2005}\)
\(=\frac{3}{4}\times\left(\frac{2006}{2005}-\frac{1}{2005}\right)\)
\(=\frac{3}{4}\times1\)
\(=\frac{3}{4}\)
b. \(\dfrac{4.2.3.5.2.6.2.5.3}{4.3.4.5.6.4.5.6}=\dfrac{1}{4}\)
c. \(\dfrac{17.2.101.2005}{17.2.101.2.2005}=\dfrac{1}{2}\)
d. \(\dfrac{2.3.4.2.11.13.17}{3.11.4.4.17.13.2}=\dfrac{1}{2}\)
a)\(\dfrac{4\times6\times10\times12\times15}{4\times12\times5\times24\times30}=\dfrac{2}{2\times2\times2}=\dfrac{2}{8}=\dfrac{1}{4}\)
b)\(\dfrac{34\times101\times2005}{17\times202\times4010}=\dfrac{2}{2\times2}=\dfrac{2}{4}=\dfrac{1}{2}\)
c)\(\dfrac{6\times8\times11\times13\times17}{33\times16\times17\times26}=\dfrac{6}{3\times2\times2}=\dfrac{1}{2}\)
Ta có:\(A=2004\times15+2004\times6-2004\)
\(=2004\times\left(15+6-1\right)\)
\(=2004\times\left(21-1\right)=2004\times20\)
\(< 20\times2005=B\)
Nên \(A< B\)
1)
\(\frac{1}{2003}\times\left(1-\frac{1}{2004}\right)\times\left(1-\frac{1}{2005}\right)\times\left(1-\frac{1}{2006}\right)\)
\(=\frac{1}{2003}\times\frac{2003}{2004}\times\frac{2004}{2005}\times\frac{2005}{2006}\)
\(=\frac{1\times2003\times2004\times2005}{2003\times2004\times2005\times2006}\)
\(=\frac{1}{2006}\)
Ta có:\(\frac{1}{2003}\times\left(1-\frac{1}{2004}\right)\times\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)
\(=\frac{1}{2003}.\frac{2003}{2004}.\frac{2004}{2005}.\frac{2005}{2006}\)
\(=\frac{1}{2006}\)
Tính nhanh mỗi biểu thức sau:
a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20
= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10
= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10
= 20 x 10 + 10
= 200 + 10
= 210
b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)
= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0
= A x 0
= 0
c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
= 0 : A
= 0
d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (30 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= (37 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)
= 0 x A
= 0
e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)
= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)
= 0 : (1 x 2 x 3 x 4 x ... x 10)
= 0 : A
= 0
g, (m : 1 - m x 1) : (m x 2008 + m x 2008)
= (m - m) : (m x 2008 + m x 2008)
= 0 : (m x 2008 + m x 2008)
= 0 : A
= 0
h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)
= (2 + 4 + 6 + 8 + m x n) x (972 - 972)
= (2 + 4 + 6 + 8 + m x n) x 0
= A x 0
= 0
l, (1 + 2 + 3 + ... + 99) x (13 x 15 - 12 x 15 - 15)
= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))
= (1 + 2 + 3 + ... + 99) x (15 x 0)
= (1 + 2 + 3 + ... + 99) x 0
= A x 0
= 0
i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)
= 0 x : (2 + 4 + 6 +...+ 98)
= 0 x A
= 0
k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)
= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))
= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)
= (0 + 1 + 2 +...+ 97 + 99) x 0
= A x 0
= 0
tra loi ca 2 phan mik tick cho