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TH1: |x-2014|^2015=1 và |x-2015|^2014=0
=>(x-2014=1 hoặc x-2014=-1) và x-2015=0
=>x=2015
TH2: |x-2014|^2015=0và |x-2015|^2014=1
=>x-2014=0 và (x-2015=1 hoặc x-2015=-1)
=>x=2014
\(\dfrac{x-3}{2014}+\dfrac{x-2}{2015}=\dfrac{x-2015}{2}+\dfrac{x-2014}{3}\)
⇔ \(\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1=\dfrac{x-2015}{2}-1+\dfrac{x-2014}{3}-1\)
⇔ \(\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2}+\dfrac{x-2017}{3}\)
⇔ \(\left(x-2017\right)\)\(\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
Do : \(\dfrac{1}{2014}< \dfrac{1}{2015}< \dfrac{1}{2}< \dfrac{1}{3}\)
⇒ \(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2}-\dfrac{1}{3}< 0\)
⇔ x - 2017 = 0
⇔ x = 2017
KL....
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 ~
Ta có:
\(M=\frac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{xyz-x^3+xyz-y^3+xyz-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{3xyz-x^3-y^3-z^3}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(-M=\frac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Xét đẳng thức phụ:
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=\left[\left(a +b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-ab\right]=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-abc-ac\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\)
Thay vào -M ta có:
\(-M=\frac{\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}=\frac{1}{2}\left(x+y+z\right)\Rightarrow M=-\frac{1}{2}\left(x+y+z\right)\)
Giờ thay: \(x=2014^{2015}-20142015;y=20142015-2015^{2014};z=2015^{2014}-2014^{2015}\)
Ta có:
\(M=-\frac{1}{2}\left(2014^{2015}-20142015+20142015-2015^{2014}+2015^{2014}-2014^{2015}\right)=0\)
(\dfrac{2-x}{2013}-1=\dfrac{1-x}{2014}-\dfrac{x}{2015}\)
\(\Leftrightarrow\dfrac{2-x+2013}{2013}-1+1=\dfrac{1-x+2014}{2014}+\dfrac{-x+2015}{2015}\) (Cộng 2 vế cho 1+1)
\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}+\dfrac{2015-x}{2015}\)
\(\Leftrightarrow\dfrac{2015-x}{2013}-\dfrac{2015-x}{2014}-\dfrac{2015-x}{2015}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow2015-x=0\) (Vì \(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\ne0\))
\(\Leftrightarrow x=2015\)
Vậy phương trình có tập nghiệm là x = 2015
\(\dfrac{2-x}{2013}-1=\dfrac{1-x}{2014}-\dfrac{x}{2015}\)
\(\Leftrightarrow\dfrac{2-x+2013}{2013}-1+1=\dfrac{1-x+2014}{2014}+\dfrac{-x+2015}{2015}\) (Cộng 2 vế cho 1+1)
\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}+\dfrac{2015-x}{2015}\)
\(\Leftrightarrow\dfrac{2015-x}{2013}-\dfrac{2015-x}{2014}-\dfrac{2015-x}{2015}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow2015-x=0\) (Vì \(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\ne0\))
\(\Leftrightarrow x=2015\)
Vậy phương trình có tập nghiệm là x = 2015
\(\dfrac{2-x}{2013}-1=\dfrac{1-x}{2014}-\dfrac{x}{2015}\)
\(\Leftrightarrow\dfrac{2-x+2013}{2013}-1+1=\dfrac{1-x+2014}{2014}+\dfrac{-x+2015}{2015}\) (Cộng 2 vế cho 1+1)
\(\Leftrightarrow\dfrac{2015-x}{2013}=\dfrac{2015-x}{2014}+\dfrac{2015-x}{2015}\)
\(\Leftrightarrow\dfrac{2015-x}{2013}-\dfrac{2015-x}{2014}-\dfrac{2015-x}{2015}=0\)
\(\Leftrightarrow\left(2015-x\right)\left(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Leftrightarrow2015-x=0\) (Vì \(\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\ne0\))
\(\Leftrightarrow x=2015\)
Vậy phương trình có tập nghiệm là x = 2015
gõ phân số ra cho mk nhìn đc ko cậu