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(2015x - 2014)³ = 8(x - 1)³ + (2013x - 2012)³

<=> 6(x - 1)(2013x - 2012)(2015x - 2014) = 0

Tới đây thì xong rồi

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....

Bài 1 : 

Ta có  : 

\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)

\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)

\(+\left(\frac{x+2013}{2011}+1\right)\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)

\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

Bài 2 : 

Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)

=> Phương trình trở thành 

\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)

\(\Rightarrow5a^2+3a-8=0\)

\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)

Vì \(a\ge0\Rightarrow a=1\)

\(\Rightarrow x^2+2x+1=1\)

\(x^2+2x=0\)

\(\Rightarrow x\left(x+2\right)=0\)

\(\Rightarrow x\in\left\{-2,0\right\}\)

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 ~

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 \(S=\left\{2015\right\}\)

Theo bài ra , ta có :

\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)

\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)

\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)

\(\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

Vậy \(x=-2016\)

Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)

Chúc bạn học tốt =)) 

\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)

\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)

\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)

\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)

\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)

\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

(x-1)/2015 + x/2014 + 1/503 - (x-3)/2013 - x/2012 - 1/1007 =0

(x-2016)/2015  + (x-2016)/2014 - (x-2016)/2012 - (x-2016)/2013 = 0

(x-2016) ( 1/2015 + 1/2016 - 1/2013 - 1/2012) = 0

Mà 1/2015 + 1/2016 - 1/2013 - 1/2012 khác 0

Suy ra x -2016=0

x=2016

Chỗ nào thắc mắc nhớ hỏi mik nhe!

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