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\(\frac{2009x2008-1}{2007x2009+2008}=\frac{2009x2007+2009-1}{2009x2007+2008}=1.\)
vậy biểu thức trên =1
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2009 + 2008 + 2007 + ..... + (x + 1) + x = 2009
x + (x + 1) + (x + 2) + .......... + 2008 + 2009 = 2009
Áp dụng công thức tính dãy số ta có :
\(\frac{\left[\left(2009-x\right):1+1\right].\left(2009+x\right)}{2}=2009\)
\(\frac{\left[2009-x+1\right]\left(2009+x\right)}{2}=2009\)
\(\left[2008-x\right]\left(2009+x\right)=4018\)
\(2008\left(2009+x\right)-x\left(2009+x\right)=4018\)
\(2008.2009+2008x-\left(2009x+x^2\right)=4018\)
2008.2009 + 2008x - 2009x - x2 = 4018
2008.2009 - x - x2 = 4018
2008.2009 - x(x + 1) = 4018
x(x + 1) = 4034072 - 4018
x(x + 1) = 4030054
Còn lại cậu dò tìm số x là được !!!
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Điều kiện: 2009.x\(\ge\)0
\(\Rightarrow\)x\(\ge\)0
\(\Rightarrow\)/x+1/, /x+2/, .... , /x+2008/\(\ge\)0
\(\Rightarrow\)/x+1/+/x+2/+...+/x+2008/=2009.x\(\Leftrightarrow\)2008x+1+2+...+2008=2009x
\(\Rightarrow\)x=2017036
Vậy \(x=2017036\)
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b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
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Dễ thấy: \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|x+2\right|\ge0\\.................\\\left|x+2008\right|\ge0\end{matrix}\right.\)\(\forall x\)
\(\Rightarrow VT=\left|x+1\right|+\left|x+2\right|+...+\left|x+2008\right|\ge0\forall x\)
\(\Rightarrow VP\ge0\forall x\Rightarrow2009x\ge0\Rightarrow x\ge0\)
Vậy \(pt\Leftrightarrow\left(x+1\right)+\left(x+2\right)+...+\left(x+2008\right)=2009x\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+2008\right)=2009x\)
\(\Leftrightarrow2008x+2017036=2009x\)
\(\Leftrightarrow2009x-2008x=2017036\Leftrightarrow x=2017036\)