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kết quả là 2008 đấy bạn
nếu nhà bạn có máy tính thì chỉ cần bấm phương trình x thì sẽ ra kết quả thôi
\(\frac{x-1}{2007}+\frac{x-2}{2006}+\frac{x-3}{2005}=\frac{x-4}{2004}+\frac{x-5}{2003}+\frac{x-6}{2002}\)
=> \(\left(\frac{x-1}{2007}-1\right)+\left(\frac{x-2}{2006}-1\right)+\left(\frac{x-3}{2005}-1\right)=\left(\frac{x-4}{2004}-1\right)+\left(\frac{x-5}{2003}-1\right)+\left(\frac{x-6}{2002}-1\right)\)
=> \(\frac{x-1+2007}{2007}+\frac{x-2+2006}{2006}+\frac{x-3+2005}{2005}=\frac{x-4+2004}{2004}+\frac{x-5+2003}{2003}+\frac{x-6+2002}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}=\frac{x-2008}{2004}+\frac{x-2008}{2003}+\frac{x-2008}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}-\frac{x-2008}{2004}-\frac{x-2008}{2003}-\frac{x-2008}{2002}=0\)
=> \(\left(x-2008\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2008 = 0 => x = 2008
Vậy x = 2008
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A=1−3+5−7+...+2001−2003+2005S=1−3+5−7+...+2001−2003+2005
=(1−3)+(5−7)+...+(2001−2003)+2005=(1−3)+(5−7)+...+(2001−2003)+2005(Có 1002 cặp)
=(−2).1002+2005=(−2).1002+2005
=−2004+2005=−2004+2005
=1
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a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
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a) 203/448>203/499>200/449
Vậy 203/448>200/449
b) Ta có:1-2002/2003=1/2003;1-2005/2006=1/2006
Vì 1/2003>1/2006 nên 2002/2003<2005/2006