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\(\frac{x-3}{2014}-\frac{x-2}{2015}=\frac{x-1}{1008}+\frac{x}{2017}-1\)

=> \(\frac{x-3}{2014}-\frac{x-2}{2015}-2=\frac{x-1}{1008}+\frac{x}{2017}-1-2\)

=> \(\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-1}{1008}-2\right)+\left(\frac{x}{2017}-1\right)\)

=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}=\frac{x-2017}{1008}+\frac{x-2017}{2017}\)

=> \(\frac{x-2017}{2014}-\frac{x-2017}{2015}-\frac{x-2017}{1008}-\frac{x-2017}{2017}\)

=> \(\left(x-2017\right)\left(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}\right)=0\)

Vì \(\frac{1}{1008}>\frac{1}{2014}\) nên \(\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}-\frac{1}{2017}< 0\)

=> x-2017=0

=>x=2017

\(\frac{x}{2012}+\frac{x-1}{2013}+\frac{x-2}{2014}-\frac{x-3}{2015}=\frac{x-4}{1008}\)

\(\Leftrightarrow\left(\frac{x}{2012}+1\right)+\left(\frac{x-1}{2013}+1\right)+\left(\frac{x-2}{2014}+1\right)-\left(\frac{x-3}{2015}+1\right)=\frac{x-4}{1008}+2\)

\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x-4+1008.2}{1008}\)

\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}=\frac{x+2012}{1008}\)

\(\Leftrightarrow\frac{x+2012}{2012}+\frac{x+2012}{2013}+\frac{x+2012}{2014}-\frac{x+2012}{2015}-\frac{x+2012}{1008}=0\)

\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{1008}\ne0\)

\(\Rightarrow x+2012=0\)\(\Leftrightarrow x=-2012\)

Vậy \(x=-2012\)

Chu Công Đứcbạn làm kết quả đúng nhưng trình bày sai rồi vế phải bạn cộng 2 nhưng vế trái bạn cộng 4???

Nhưng nếu vế phải +2 thì kết quả có thay đổi ko vậy ạ

b)       \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt   \(x^2+3x=t\)   ta có:

          \(t\left(t+2\right)-24=0\)

\(\Leftrightarrow\)\(t^2+2t-24=0\)

\(\Leftrightarrow\)\(\left(1-4\right)\left(1+6\right)=0\)

đến đây bn giải tiếp

PT đã cho tương đương với:

\(\left(\frac{x}{2017}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+3}{2014}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}\right)=\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2014}\right)\)

\(\Leftrightarrow x+2017=0\Leftrightarrow x=-2017\)

Ta có: \(\frac{x}{2017}+\frac{x+1}{2016}=\frac{x+2}{2015}+\frac{x+3}{2014}\)

\(\Leftrightarrow\frac{x}{2017}+1+\frac{x+1}{2016}+1=\frac{x+2}{2015}+1+\frac{x+3}{2014}+1\)

\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}=\frac{x+2017}{2015}+\frac{x+2017}{2014}\)

\(\Leftrightarrow\frac{x+2017}{2017}+\frac{x+2017}{2016}-\frac{x+2017}{2015}-\frac{x+2017}{2014}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\right)=0\)

mà \(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\)

nên x+2017=0

hay x=-2017

Vậy: S={-2017}

(x+2/2014)+1 + (x+1/2015)+1 = (x+2016)+1 + (x-1/2017)+1

(x+2016/2014) + (x+2016/2015) - (x+2016/2016) - (x-2016/2017)=0

=>(x+2016)(1/2014+1/2015-1/2016-1/2017)

vì 1/2014+1/2015-1/2016-1/2017 luôn khác 0 => x+2016=0

=> x=-2016

Đặt \(\sqrt{x-2014}=a;\sqrt{y-2015}=b;\sqrt{z=2016}=c\)(với a,b,c>0). Khi đó pt trở thành: 

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)\(\Leftrightarrow\left(\frac{1}{4}-\frac{1}{a}+\frac{1}{a^2}\right)+\left(\frac{1}{4}-\frac{1}{b}+\frac{1}{b^2}\right)+\left(\frac{1}{4}-\frac{1}{c}+\frac{1}{c^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{2}-\frac{1}{a}\right)^2+\left(\frac{1}{2}-\frac{1}{b}\right)^2+\left(\frac{1}{2}-\frac{1}{c}\right)^2=0\Leftrightarrow a=b=c=2\)

\(\Rightarrow x=2018;y=2019;z=2020\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}-\left(\frac{1}{x-2014+y-2015+z-2016}\right)=\frac{3}{4}\)

\(\frac{\sqrt{x-2014}}{x-2014}+\frac{\sqrt{y-2015}}{y-2015}+\frac{\sqrt{z-2016}}{z-2016}+0=\frac{3}{4}\)

\(\frac{\sqrt{x}-\sqrt{2014}}{x-2014}+\frac{\sqrt{y}-\sqrt{2015}}{y-2015}+\frac{\sqrt{z}-\sqrt{2016}}{z-2016}=\frac{3}{4}\)

\(x=2018,y=2019,z=2020\)

ĐK : \(\hept{\begin{cases}x>2014\\y>2015\\z>2016\end{cases}}\)

\(\frac{\sqrt{x-2014}-1}{x-2014}+\frac{\sqrt{y-2015}-1}{y-2015}+\frac{\sqrt{z-2016}-1}{z-2016}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}-\frac{\sqrt{x-2014}-1}{x-2014}+\frac{1}{4}-\frac{\sqrt{y-2015}-1}{y-2015}+\frac{1}{4}-\frac{\sqrt{z-2016}-1}{z-2016}=0\)

\(\Leftrightarrow\frac{x-2010-4\sqrt{x-2014}}{4\left(x-2014\right)}+\frac{y-2011-4\sqrt{y-2015}}{4\left(y-2015\right)}+\frac{z-2012-4\sqrt{z-2016}}{4\left(x-2014\right)}=0\)

\(\Leftrightarrow\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}+\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}+\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}=0\)( 1 )

Mà \(\hept{\begin{cases}\frac{\left(2-\sqrt{x-2014}\right)^2}{4\left(x-2014\right)}\ge0\forall x>2014\\\frac{\left(2-\sqrt{y-2015}\right)^2}{4\left(y-2015\right)}\ge0\forall y>2015\\\frac{\left(2-\sqrt{z-2016}\right)^2}{4\left(z-2016\right)}\ge0\forall z>2016\end{cases}}\)( 2 )

Từ ( 1 ) và ( 2 ) => \(\hept{\begin{cases}\left(2-\sqrt{x-2014}\right)^2=0\\\left(2-\sqrt{y-2015}\right)^2=0\\\left(2-\sqrt{z-2016}\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}\sqrt{x-2014}=2\\\sqrt{y-2015}=2\\\sqrt{z-2016}=2\end{cases}}\)<=>\(\hept{\begin{cases}x=2018\\y=2019\\z=2020\end{cases}}\)( tmđk )

Vậy ( x ; y ; z ) = ( 2018 ; 2019 ; 2020 )