a Tìm x , biết : 1\(\frac{3}{5}\) + [ \(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\)]  x = \(\frac{16}{5}\) 

b Chứng minh rằng số tự nhiên A chia hết cho 2009 , với 

A =   1 . 2 .3 ... 2007 . 2008 ( 1 + \(\frac{1}{2}\) + ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\))

                                                                           Giải

a 1\(\frac{3}{5}\)+ (\(\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{37}}{\frac{5}{7}+\frac{5}{17}+\frac{5}{37}}\)) x = \(\frac{16}{5}\)\(\Leftrightarrow\) \(\frac{8}{5}\)+ [\(\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{37}\right)}\)x = \(\frac{16}{5}\)

\(\Leftrightarrow\)\(\frac{8}{5}\) + \(\frac{2}{5}\)x = \(\frac{16}{5}\)\(\Leftrightarrow\)\(\frac{2}{5}\)x = \(\frac{16}{5}\)\(-\)\(\frac{8}{5}\) \(\Leftrightarrow\) x = \(\frac{2}{5}\)x \(\Leftrightarrow\)= \(\frac{8}{5}\) : \(\frac{2}{5}\)\(\Leftrightarrow\)x=4

b 1 + \(\frac{1}{2}\)+ \(\frac{1}{3}\)+ ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\) 

 = (1 + \(\frac{1}{2008}\))  + (\(\frac{1}{2}\)+ \(\frac{1}{2007}\)) + ... + (\(\frac{1}{2004}\)+ \(\frac{1}{2005}\)) 

= (1 + \(\frac{1}{2008}\)) + (\(\frac{1}{2}\)+ \(\frac{1}{2007}\)) + ... + (\(\frac{1}{1004}\)+ \(\frac{1}{1005}\))

= \(\frac{2009}{1\times2008}\) + \(\frac{2009}{2\times2007}\) +  ... + \(\frac{2009}{1004\times1009}\) 

= 2009(\(\frac{1}{1\times2008}\) + \(\frac{1}{2\times2007}\)+ ... + \(\frac{1}{1004\times1005}\)

Do đó A = 1 . 2 .3 ... 2007 . 2008 . (1 + \(\frac{1}{2}\) + \(\frac{1}{3}\) + ... + \(\frac{1}{2007}\)+ \(\frac{1}{2008}\))

             = 2009(1 . 2 . 3 ... 2007 . 2008 (\(\frac{1}{1.2008}\) + \(\frac{1}{2.2007}\)+ ... + \(\frac{1}{1004.1005}\) ) \(⋮\) 2009

Vì 1 . 2 . 3 ... 1007 . 2008 (\(\frac{1}{1.2008}\) + \(\frac{1}{2.2007}\) + ... + \(\frac{1}{2004.2005}\)) là một số tự nhiên 

CÁC BẠN CÓ AI GIỐNG CÁCH LÀM CỦA MÌNH THÌ TRẢ LỜI NHÉ

Tìm x biết:   \(\frac{3}{2x+1}\)+ \(\frac{10}{4x+2}\)- \(\frac{6}{6x+3}\)=\(\frac{12}{26}\)

Chứng minh rằng:      \(\frac{1}{5^2}\)+ \(\frac{1}{6^2}\)+...+ \(\frac{1}{2007^2}\)> \(\frac{1}{5}\)

rút gọn:  M = ( 1+\(\frac{1}{3}\)+ \(\frac{1}{5}\)+ ... + \(\frac{1}{999}\)) : (\(\frac{1}{1.999}\)+ \(\frac{1}{3.997}\)+...+\(\frac{1}{997.3}\)+\(\frac{1}{999.1}\))

Giúp mình nha!!!

cho B=\(\frac{1}{4^2}\)+ \(\frac{1}{6^2}\)+ \(\frac{1}{8^2}\)+....+\(\frac{1}{2006^2}\). CM: B<\(\frac{334}{2007}\)

Cho C= \(\frac{8}{9}\)+ \(\frac{24}{25}\)+ \(\frac{48}{49}\)+....+ \(\frac{200.202}{201^2}\). CM: C>99,75

a,{3849-2.[146-(2.14+6:2)]}.32

b,134-2.{156-6[54-2.(9+6)]}

c,\(\frac{5}{7}\).\(\frac{12}{11}\)+\(\frac{5}{7}\).\(\frac{17}{11}\)-\(\frac{5}{7}\).\(\frac{18}{11}\)+2\(\frac{1}{3}\)

d,\(\frac{2}{3}\)+\(\frac{3}{5}\)+\(\frac{5}{3}\)-\(\frac{1}{5}\)+1

e,\(\frac{1}{6}\)+\(\frac{1}{12}\)+\(\frac{1}{20}\)+\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)(Lưu ý chỉ phân số \(\frac{1}{72}\)mới có thể quy đồng)

cm 

a,A=1+\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+....+\(\frac{1}{100^2}\)<2

b,B=1+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+....+\(\frac{1}{63}\)<6

c,C=\(\frac{1}{2}\).\(\frac{3}{4}\).\(\frac{5}{6}\).\(\frac{7}{8}\)...\(\frac{9999}{10000}\)<\(\frac{1}{100}\)

bài 1

a,\(\frac{x}{5}\)=\(\frac{2}{3}\)

b,\(\frac{x}{3}\)-\(\frac{1}{2}\)=\(\frac{1}{5}\)

c,\(\frac{x}{5}\)+\(\frac{1}{2}\)=\(\frac{6}{10}\)

d,\(\frac{x+3}{15}\)=\(\frac{1}{3}\)

e,\(\frac{x-12}{4}\)=\(\frac{1}{2}\)

h,\(\frac{4}{5}\)+x=\(\frac{2}{3}\)

i,\(\frac{3}{4}\)-x=\(\frac{1}{3}\)

k,\(\frac{-5}{6}\)-x=\(\frac{2}{3}\)

l,x-\(\frac{5}{9}\)=\(\frac{-2}{3}\)

m,\(\frac{1}{2}\)x+\(\frac{1}{2}\)=\(\frac{5}{2}\)

n,\(\frac{-2}{3}\)-\(\frac{1}{3}\)(2x-5)=\(\frac{3}{2}\)

p,(3x-1)(\(-\frac{1}{2}\)x+5)=0

q,\(3\frac{1}{3}\)+\(\frac{5}{6}\)x=\(3\frac{1}{2}\)

s,\(\frac{1}{2}\)-\(\frac{2}{3}\)x=\(\frac{7}{12}\)

t,\(\frac{3}{4}\)x+\(\frac{1}{5}\)=\(\frac{1}{6}\)

u,\(\frac{3}{8}\)-\(\frac{1}{6}\)x=\(\frac{1}{4}\)

mn ơi'-' xin hãy giúp mk với ạ>^<,mk cần rất gấp,mai nộp r ạ,mong mn giúp mk với ạ:(( huhuhu r mk tik cho:<<

Tính a) (4 \(\frac{5}{37}\)- 3\(\frac{4}{5}\)+8\(\frac{15}{29}\)) - (3\(\frac{5}{37}\)- 6\(\frac{14}{29}\)

b) 60 \(\frac{7}{13}\)x (8\(\frac{7}{10}\)) +50\(\frac{8}{13}\)x (-8 \(\frac{7}{10}\)) - 11 \(\frac{2}{13}\)x (-8 \(\frac{7}{10}\)

c) (2 \(\frac{5}{6}\)+ 1\(\frac{4}{9}\)) : ( 10 \(\frac{1}{2}\)- 9 \(\frac{1}{2}\)

d) 1 \(\frac{5}{18}\)- \(\frac{5}{18}\)x (\(\frac{1}{15}\)+1\(\frac{1}{12}\))

e)\(\frac{-1}{7}\)x (9 \(\frac{1}{2}\) - 8,75) :\(\frac{2}{7}\) +0,625 :1 \(\frac{2}{3}\)

BT1; Tính;

a,2:\(\frac{-1}{3}\)+\(\frac{1}{3}\).\(\frac{2}{7}\)+\(\frac{1}{3}\).\(\frac{5}{7}\)

b,9,12+88%-1\(\frac{1}{5}\):\(\frac{1}{10}\)

c,\(\frac{5}{6}\)x+\(\frac{1}{4}\)x=\(\frac{1}{6}\)-\(\frac{5}{4}\)