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Bạn tham khảo nhé :
Ta có :
\(\frac{x-3}{2012}+\frac{x-2}{2013}=\frac{x-2013}{2}+\frac{x-2012}{3}\)
\(\Leftrightarrow\)\(\left(\frac{x-3}{2012}-1\right)+\left(\frac{x-2}{2013}-1\right)=\left(\frac{x-2013}{2}-1\right)+\left(\frac{x-2012}{3}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}=\frac{x-2015}{2}+\frac{x-2015}{3}\)
\(\Leftrightarrow\)\(\frac{x-2015}{2012}+\frac{x-2015}{2013}-\frac{x-2015}{2}-\frac{x-2015}{3}=0\)
\(\Leftrightarrow\)\(\left(x-2015\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2}-\frac{1}{3}\ne0\)
\(\Rightarrow\)\(x-2015=0\)
\(\Rightarrow\)\(x=2015\)
Vậy \(x=2015\)
Chsuc bạn học tốt ~
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)
pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
Chúc bạn học tốt ~
\(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
Đặt \(\hept{\begin{cases}2x^2+x-2013=m\\x^2-5x-2012=n\end{cases}}\)nên ta có phương trình:
\(m^2+4n^2=4nm\)
\(\Leftrightarrow m^2-2.m.2n+\left(2n\right)^2=0\)
\(\Leftrightarrow\left(m-2n\right)^2=0\)
Tự làm nốt...
Bạn học trường nào thế?
\(\dfrac{x-3}{2012}+\dfrac{x-2}{2013}=\dfrac{x-2013}{2}+\dfrac{x-2012}{3}\)(mk nghĩ đề như thế này)
\(\Leftrightarrow\dfrac{x-3}{2012}-1+\dfrac{x-2}{2013}-1=\dfrac{x-2013}{2}-1+\dfrac{x-2012}{3}-1\)
\(\Leftrightarrow\dfrac{x-2015}{2012}+\dfrac{x-2015}{2013}=\dfrac{x-2015}{2}+\dfrac{x-2015}{3}\)
\(\Leftrightarrow\left(x-2015\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x=2015\)(vì \(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))
\(\dfrac{x-3}{2012}+\dfrac{x-2}{2013}=\dfrac{x-2013}{2}+\dfrac{x-2015}{3}\\ \Leftrightarrow\left(\dfrac{x-3}{2012}-1\right)+\left(\dfrac{x-2}{2013}-1\right)=\left(\dfrac{x-2013}{2}-1\right)+\left(\dfrac{x-2015}{3}-1\right)\\ \Leftrightarrow\dfrac{x-2018}{2012}+\dfrac{x-2018}{2013}-\dfrac{x-2018}{2}-\dfrac{x-2018}{3}=0\\ \Leftrightarrow\left(x-2018\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \Leftrightarrow x-2018=0\left(\text{Vì }\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\right)\\ x=2018\)
Vậy phương trình có nghiệm \(x=2018\)
(2x2+x-2013)2+4 (x2-5x-2012)2= 4 (2x2+x-2013)(x2-5x-2012)
Dat \(\hept{\begin{cases}a=2x^2+x-2013\\b=x^2-5x-2012\end{cases}}\)ta co phuong trinh
(2x2+x-2013)2+4 (x2-5x-2012)2= 4 (2x2+x-2013)(x2-5x-2012)
<=>\(a^2+4b^2=4ab\)
<=>\(a^2+4b^2-4ab=0\)
<=>\(\left(a-2b\right)^2=0\)
<=>\(a=2b\)
=>\(2x^2+x-2013=2x^2-10x-4024\)
<=>\(11x=2011\)
<=>x=\(\frac{2011}{11}\)
