1+3+5+7+....+1993+1995+2xX =996006
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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
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
\(\frac{3}{5}+\frac{6}{11}+\frac{7}{13}+\frac{2}{5}+\frac{16}{11}+\frac{19}{13}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{6}{11}\frac{16}{11}\right)+\left(\frac{7}{13}+\frac{19}{13}\right)\)
\(=1+2+2\)
\(=5\)
a. 3/5 + 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.
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
(1+3+5+7+...+1995)+2x=996006
[(1995+1).998 : 2] + 2x = 996006
996004+2x=996006
=> 2x = 2
=> x=1
(1+3+5+7+...+1995)+2x=996006
[(1995+1).998 : 2] + 2x = 996006
996004+2x=996006
=> 2x = 2
=> x=1