tim x biết
/x-2010/+/x-2011/=2012
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suy ra hai truong hop
1 : x-2011=x-2012
suy ra x-x=2011-2012(loai)
2 : x-2011=-(x-2012)
suy ra : x-2011=-x+2012
2x=2011+2012
2x=4023
x=2011.5
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a/ \(\left|x-2011\right|=x-2012\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2011=x-2012\\x-2011=-x+2012\end{matrix}\right.\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x-x=-2012+2011\\x+x=2012+2011\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}0x=-1\left(loại\right)\\2x=4023\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{4023}{2}\)
Vậy ...
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
Ta có: \(\left|x-2010\right|+\left|x-2011\right|=2012\)
Nếu \(x\le2010\Rightarrow2010-x+2011-x=2012\Rightarrow x=\frac{2009}{2}\) (thỏa mãn)
Nếu \(2010< x< 2011\Rightarrow x-2010+2011-x=2012\Rightarrow1=2012\) (không thỏa mãn)
Nếu \(x\ge2011\Rightarrow x-2010+x-2011=2012\Rightarrow x=\frac{6033}{2}\) (thỏa mãn)
Vậy \(x=\left\{\frac{2009}{2};\frac{6033}{2}\right\}\)
/x-2010/+/x-2011/=2012 (1)
Nếu x \(\le\) 2010 từ (1) suy ra : 2010 - x + 2011 - x = 2012 => x = 2009/2 (TM)
Nếu 2010 < x < 2011 từ (1) suy ra : x - 2010 + 2011 - x = 2012 => 1 = 2012 (loại)
Nếu x \(\ge\) 2011 từ (1) suy ra : x - 2010 + x - 2011 = 2012 => x = 6033/2 (TM)
Vậy...