bn bị rảnh ak ?

ko trả lời thì đừng có viết linh tinh

\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}+\frac{x+2045}{10}=0\)

\(\Leftrightarrow\frac{x+1}{2014}+1+\frac{x+2}{2013}+1+\frac{x+3}{2012}+1+\frac{x+2045}{10}-3=0\)

\(\Leftrightarrow\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}+\frac{x+2045-3.10}{10}=0\)

\(\Leftrightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}+\frac{x+2015}{10}=0\)

\(\Leftrightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{10}\ne0\)

Nên x + 2015 = 0 <=> x = -2015

Vậy x = -2015

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

\(x+\dfrac{2015}{2}+x+\dfrac{2013}{3}>-x-\dfrac{2013}{-4}-x-\dfrac{2012}{-5}\\ \Leftrightarrow x+1007.5+x+\dfrac{2014}{3}>\dfrac{2013x}{4}-x+402.4\\ \Leftrightarrow x+\dfrac{10075}{10}+x+\dfrac{2014}{3}>\dfrac{2013x}{4}-x+\dfrac{4024}{10}\\ \Leftrightarrow2x+\dfrac{2015}{2}+\dfrac{2014}{3}>\dfrac{2013x-4x}{4}+\dfrac{4024}{10}\\ \Leftrightarrow2x+\dfrac{6045+4028}{6}>\dfrac{2009x}{4}+\dfrac{4024}{10}\)

\(\Leftrightarrow2x+\dfrac{10073}{6}>\dfrac{2009x}{4}+\dfrac{4024}{10}\\ \Leftrightarrow120x+10.10073>15.2009x+6.4024\\ \Leftrightarrow120x+100730>30135x+24144\\ \Leftrightarrow-30015x>-76586\\ \Leftrightarrow x< \dfrac{76586}{30015}\)

Tập nghiệm \(S=\left\{x|x< \dfrac{76586}{30015}\right\}\)

x^2+x*y-2012*x-2013*y-2014=0

<=> x^2+x*y+x-2013*x-2013*y-2014=0

<=>x*(x+y+1)-2013*(x+y+1)=0

<=> (x-2013)*(x+y+1)=1

do x,y nguyen nen

(x,y) la (2014;-2014);(2012;-2014)

bạn là nam hay nữ zở

bn nhìn tên rồi đoán nha bn

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 ~