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đặt 2009-x=a,x-2010=b
suy ra a^2+ab+b^2/a^2-ab+b^2=19/49
suy ra 49(a^2+ab+b^2)=19(a^2-ab+b^2)
49a^2+49ab+49b^2=19a^2-19ab+19b^2
30a^2+68ab+30b^2=0
30a^2+50ab+18ab+30b^2=0
10a(3a+5b)+6b(3a+5b)=0
(3a+5b)(10a+6b)=0
suy ra 3a+5b=0 hoặc 10a+6b=0
thế vào lại rồi tìm x
Đặt \(2009-x=t\Rightarrow x-2010=-\left(2009-x\right)-1=-t-1\)
Suy ra:
\(\frac{t^2+t\left(-t-1\right)+\left(-t-1\right)^2}{t^2-t\left(-t-1\right)+\left(-t-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2-t\left(t+1\right)+\left(t+1\right)^2}{t^2+t\left(t+1\right)+\left(t+1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2-t^2-t+t^2+2t+1}{t^2+t^2+t+t^2+2t+1}=\frac{19}{49}\)
\(\Leftrightarrow\frac{t^2+t+1}{3t^2+3t+1}=\frac{19}{49}\)
\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)
\(\Leftrightarrow8t^2+8t-30=0\)
\(\Leftrightarrow4t^2+4t-15=0\Leftrightarrow4t^2+4t+1=16\)
\(\Leftrightarrow\left(2t+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}2t+1=-4\\2t+1=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2t=-5\\2t=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\frac{5}{2}\\t=\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2009-x=-\frac{5}{2}\\2009-x=\frac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4023}{2}\\x=\frac{4015}{2}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{4015}{2};\frac{4023}{2}\right\}\)
Lời giải của mình ở đây nhé bạn!
http://olm.vn/hoi-dap/question/424173.html
Đặt \(\left\{{}\begin{matrix}x-2010=a\\2009-x=b\end{matrix}\right.\)
Theo đề bài ta có:
\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{b^2+ab+a^2}{b^2-ab+a^2}=\dfrac{19}{49}\)
\(\Leftrightarrow19\left(b^2-ab+a^2\right)=49\left(b^2+ab+a^2\right)\)
\(\Leftrightarrow19b^2-19ab+19a^2-49b^2-49ab-49a^2=0\)
\(\Leftrightarrow-30a^2-68ab-30b^2=0\)
\(\Leftrightarrow-2\left(15a^2+34ab+15b^2\right)=0\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow15a^2+25ab+9ab+15b^2=0\)
\(\Leftrightarrow5a\left(3a+5b\right)+3b\left(3a+5b\right)=0\)
\(\Leftrightarrow\left(3a+5b\right)\left(5a+3b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3a+5b=0\\5a+3b=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\left(x-2010\right)+5\left(2009-x\right)=0\\5\left(x-2010\right)+3\left(2009-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-6030+10045-5x=0\\5x-10050+6027-3x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x+4015=0\\2x-4023=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4015\\2x=4023\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4015}{-2}=2007,5\\x=\dfrac{4023}{2}=2011,5\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=2007,5\\x=2011,5\end{matrix}\right.\)
Đặt a=(2009-x)2
b=(x-2010)2
Theo đề bài ta có
\(\dfrac{\text{a^2+ab+b^2}}{a^2-ab+b^2}=\dfrac{19}{49}\)
\(\text{49(a^2+ab+b^2)}=19\left(a^2-ab+b^2\right)\)
\(\text{30a^2+68ab+30b^2=0}\)
\(\text{15a^2+34ab+15b^2=0}\)
\(\text{15a^2+9ab+25ab+15b^2=0}\)
\(\text{3a(5a+3b)+5(3b+5a)=0}\)
\(\text{(5a+3b)(3a+5b)=0}\)
\(\left[{}\begin{matrix}3a+5b=0\\3b+5a=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}3\left(2009-x\right)=5\left(x-2010\right)\\5\left(2009-x\right)=3\left(x-2010\right)\end{matrix}\right.\)
\(-8x=-6030-10045\) hay \(8x=-10050-6027\)
\(x\simeq2009\),375 hay \(x\simeq2009,625\)
Đặt x-2009=a\(\Leftrightarrow\dfrac{\left(x-2009\right)^2-\left(x-2009\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(x-2009\right)^2+\left(x-2009\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{a^2-a\left(a-1\right)+\left(a-1\right)^2}{a^2+a\left(a-1\right)+\left(a-1\right)^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{a^2-a^2+a+a^2-2a+1}{a^2+a^2-a+a^2-2a+1}=\dfrac{19}{49}\)
=>\(\dfrac{a^2-a+1}{3a^2-3a+1}=\dfrac{19}{49}\)
=>49a^2-49a+49-57a^2+57a-19=0
=>-8a^2+8a+30=0
=>a=5/2 hoặc a=-3/2
=>x-2009=5/2 hoặc x-2009=-3/2
=>x=4023/2 hoặc x=4015/2
đề sai rồi th ngu