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−2010=a2009−x=b{x−2010=a2009−x=b
Theo đề bài ta có:
(2009−x)2+(2009−x)(x−2010)+(x−2010)2(2009−x)2−(2009−x)(x−2010)+(x−2010)2=1949(2009−x)2+(2009−x)(x−2010)+(x−2010)2(2009−x)2−(2009−x)(x−2010)+(x−2010)2=1949
⇔b2+ab+a2b2−ab+a2=1949⇔b2+ab+a2b2−ab+a2=1949
⇔19(b2−ab+a2)=49(b2+ab+a2)⇔19(b2−ab+a2)=49(b2+ab+a2)
⇔19b2−19ab+19a2−49b2−49ab−49a2=0⇔19b2−19ab+19a2−49b2−49ab−49a2=0
⇔−30a2−68ab−30b2=0⇔−30a2−68ab−30b2=0
⇔−2(15a2+34ab+15b2)=0⇔−2(15a2+34ab+15b2)=0
⇔15a2+34ab+15b2=0⇔15a2+34ab+15b2=0
⇔15a2+25ab+9ab+15b2=0⇔15a2+25ab+9ab+15b2=0
⇔5a(3a+5b)+3b(3a+5b)=0⇔5a(3a+5b)+3b(3a+5b)=0
⇔(3a+5b)(5a+3b)=0⇔(3a+5b)(5a+3b)=0
⇔[3a+5b=05a+3b=0⇔[3a+5b=05a+3b=0
⇔[3(x−2010)+5(2009−x)=05(x−2010)+3(2009−x)=0⇔[3(x−2010)+5(2009−x)=05(x−2010)+3(2009−x)=0
⇔[3x−6030+10045−5x=05x−10050+6027−3x=0⇔[3x−6030+10045−5x=05x−10050+6027−3x=0
⇔[−2x+4015=02x−4023=0⇔[−2x=−40152x=4023⇔[−2x+4015=02x−4023=0⇔[−2x=−40152x=4023
⇔⎡⎢ ⎢⎣x=−4015−2=2007,5x=40232=2011,5⇔[x=−4015−2=2007,5x=40232=2011,5
Vậy [x=2007,5x=2011,5
Đặt a=(2009-x)2
b=(x-2010)2
Theo đề bài ta có
a^2+ab+b^2a2−ab+b2=1949a^2+ab+b^2a2−ab+b2=1949
49(a^2+ab+b^2)=19(a2−ab+b2)49(a^2+ab+b^2)=19(a2−ab+b2)
30a^2+68ab+30b^2=030a^2+68ab+30b^2=0
15a^2+34ab+15b^2=015a^2+34ab+15b^2=0
15a^2+9ab+25ab+15b^2=015a^2+9ab+25ab+15b^2=0
3a(5a+3b)+5(3b+5a)=03a(5a+3b)+5(3b+5a)=0
(5a+3b)(3a+5b)=0(5a+3b)(3a+5b)=0
[3a+5b=03b+5a=0[3a+5b=03b+5a=0
[3(2009−x)=5(x−2010)5(2009−x)=3(x−2010)[3(2009−x)=5(x−2010)5(2009−x)=3(x−2010)
−8x=−6030−10045−8x=−6030−10045 hay 8x=−10050−60278x=−10050−6027
x≃2009x≃2009,375 hay x≃2009,625
Đặt a=2009−xa=2009−x, b=x−2010b=x−2010 ⇒a+b=−1⇒a+b=−1.
Ta có : 1949=a2+ab+b2a2−ab+b2=(a+b)2−ab(a+b)2−3ab=1−ab1−3ab1949=a2+ab+b2a2−ab+b2=(a+b)2−ab(a+b)2−3ab=1−ab1−3ab ⇒ab=−334⇒ab=−334.
Từ đó tính được aa hoặc bb, suy ra xx.