\(|x-2009|+|2009-x|=2\)
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\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\left(1\right)\)
\(Đkxđ:x\ne2009;x\ne2010\)
Đặt \(t=x-2010\left(t\ne0\right)\)
\(\Rightarrow2009-x=-\left(t+1\right)\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(t+1\right)^2-\left(t+1\right)t+t^2}{\left(t+1\right)^2+\left(t+1\right)t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+2t+1-t^2-t+t^2}{t^2+2t+1+t^2+t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+t+1}{3t^2+3t+1}=\dfrac{19}{49}\)
\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)
\(\Leftrightarrow8t^2+8t-30=0\)
\(\Leftrightarrow4t^2+4t-15=0\)
\(\Leftrightarrow\left(4t^2+4t+1\right)-16=0\)
\(\Leftrightarrow\left(2t+1\right)^2=16=4^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=4\\2t+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3}{2}\\t=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2010=\dfrac{3}{2}\\x-2010=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4023}{2}\\x=\dfrac{4015}{2}\end{matrix}\right.\)
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dat a =2009-x
b=x-2010
ta co : a^2+ab+b^2/a^2-ab+b^2 =19/49
<=>49a^2+49ab+49b^2=19a^2-19a+19b^2
<=>30a^2+68a+30b^2=0
<=>15a^2+34ab+15b^2=0
<=>15a^2+9ab+25ab+15b^2=0
<=>3a(5a+3b)+5b(5a+3b)=0
<=>(5a+3b)(3a+5b)=0
<=>5a+3b=0 hoac 3a+5b=0
vs 5a +3b=0 <=>5(2009-x)+3(x-2010)=0=>x=......
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Lời giải của mình ở đây nhé!
http://olm.vn/hoi-dap/question/424173.html
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2009 x 7 + 2009 x 2 + 2009
= 2009 x (7 + 2) + 2009
= 2009 x 9 + 2009
= 18081 + 2009
= 20090
hoặc : 2009 x 7 + 2009 x 2 + 2009
= 2009 x (7 + 2 + 1)
= 2009 x 10
= 20090
x=2010
\(|x-2009|+|2009-x|=2\)
\(\Leftrightarrow|x-2009|+|x-2009|=2\)
\(\Leftrightarrow2.|x-2009|=2\)\(\Leftrightarrow|x-2009|=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2009=-1\\x-2009=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2008\\x=2010\end{cases}}\)
Vậy \(x=2008\)hoặc \(x=2010\)