\(\frac{78}{56}+\frac{136}{52}=\frac{39}{28}+\frac{34}{13}=\frac{ }{ }\)\(\frac{507}{364}+\frac{952}{364}=\frac{507+952}{364}=\frac{1459}{364}\)

2014 x (128 x 68 - 64 x 136) + 2015

2014 x (8704 - 8704) + 2015

2014 x 0 + 2015

0 + 2015

= 2015

A = 2015

3

#@Hoàng Minh Phúc, bn giải chi tiết ra nhé!

a, \(\frac{17}{14}+4=\frac{17}{14}+\frac{56}{14}=\frac{73}{14}\)

b, \(\frac{-17}{14}+4=\frac{-17}{14}+\frac{56}{14}=\frac{39}{14}\)

12/17+19/17=31/17

31/17 bạn nhé

a,\(\frac{26}{7}\)

b,\(\frac{78}{19}\)

c,\(\frac{13}{11}\)

d,\(1\)

a, \(4-\frac{2}{7}\)=\(\frac{26}{7}\)

b,5-\(\frac{17}{19}=\frac{78}{19}\)

c,\(\frac{35}{11}-2=\frac{13}{11}\)

d,\(\frac{68}{17}-3=1\)

k mình nha

\(=\frac{15}{17}x\left(\frac{45}{13}-\frac{10}{13}-\frac{9}{13}-1\right)\\ =\frac{15}{17}x1=\frac{15}{17}\)

\(\frac{15}{17}\times\frac{45}{13}-\frac{15}{17}\times\frac{10}{13}-\frac{9}{13}\times\frac{15}{17}-\frac{15}{17}\)

\(=\frac{15}{17}\times\left(\frac{45}{13}-\frac{10}{13}-\frac{9}{13}-1\right)\)

\(=\frac{15}{17}\times1\)

\(=\frac{15}{17}\)

Trả lời

Sửa đề câu a 1 tí

a)39/16:5/8-17/16:5/8+4/5

=(39/16-17/16):5/8+4/5

=11/8:5/8+4/5

=11/5+4/5

=15/5=3

b)16/17.12/19+16/17.7/19-2/3

=16/17.(12/19+7/19)-2/3

=16/17-2/3

=48/56-14/56

=34/56=17/28

Học tốt !

Ta có:

\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)

\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)

Ta lại có: 

\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)

\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)

\(=\frac{1.2.3...36}{1.2.3...36}=1\)

Từ đây ta suy ra được

\(A-B=1-1=0\)

BAN  CO THE TINH RO BIEU THUC B KO?

a, vì 5/7<1<15/12

nên 5/7<15/12

b,17/21<17/19

\(\frac{5}{7}\)<\(\frac{15}{12}\)

\(\frac{17}{21}\)<\(\frac{17}{19}\)