Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)

\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)

\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)

\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)

\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)

\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)

Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)

=> x - 2000 = 0 

=> x = 2000

a. 3/5 + 6/11 + 7/13 + 2/5 + 16/11 + 19/13

 = ( 3/5 + 2/5 ) + ( 6/11 + 16/11 ) + ( 7/13 + 19/13)

 = 1 + 2 + 2

 = 5.

de om 5 chu may

vì 5/5+22/11+26/13=1+2+2=5

a) \(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}\)

\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)

\(=1+2+2=5\)

b) \(\frac{1995}{1997}x\frac{1990}{1993}x\frac{1997}{1994}x\frac{1993}{1995}x\frac{997}{995}=\frac{1995x1990x1997x1993x997}{1997x1993x1994x1995x995}=\frac{1990x997}{1994x995}=\frac{995x2x997}{997x2x995}=1\)

a) =(3/5+2/5)+(6/11+16/11)+(19/13+7/13)

=5/5+22/11+26/13

=1+2+3

=6

a. 

(1 + 3 + 5 + ... + 2007 + 2009 + 2011) x (125125 x 127 - 127127 x 125)

= (1 + 3 + 5 + ... + 2007 + 2009 + 2011) x 0

= 0

b.

\(\frac{2006\times125+1000}{126\times2006-1006}=\frac{2006\times125+1000}{125\times2006+1\times2006-1006}=\frac{2006\times125+1000}{125\times2006+1000}=1\)

a, 

( 1+3+5+7+…+2003+2005) x (125 125 x 127 – 127 127 x 125)

= ( 1+3+5+7+…+2003+2005) x (125 x 1001 x 127 – 127 x 1001x 125)

=  ( 1+3+5+7+…+2003+2005) x                   0  =    0

A = 0

B= 2006

C= 1993

Tui làm sai á ^^

\(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}\)

\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)

\(=1+2+2\)

\(=5\)

a

        [3/5+2/5]+[7/13+19/13]+[6/11+16/11]=5