Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz-xyz=0\)

\(\Leftrightarrow\left(x^2y+xy^2\right)+\left(yz^2+z^2x\right)+\left(zx^2+2xyz+y^2z\right)=0\)

\(\Leftrightarrow xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(xy+z^2+yz+zx\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

=> x = -y hoặc y = -z hoặc z = -x

Không mất tổng quát giả sử x = -y, khi đó:

\(\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=-\frac{1}{y^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{z^{2015}}\)

\(\frac{1}{x^{2015}+y^{2015}+z^{2015}}=\frac{1}{-y^{2015}+y^{2015}+z^{2015}}=\frac{1}{z^{2015}}\)

\(\Rightarrow\frac{1}{x^{2015}}+\frac{1}{y^{2015}}+\frac{1}{z^{2015}}=\frac{1}{x^{2015}+y^{2015}+z^{2015}}\)

câu 2  :

ab+  bc + ca = 2015 

=> 2015 +a^2 = a^2 + ab + bc + ca 

=> 2015 + a^2 = a(a+b ) + c( a + b ) = ( a + c )( a + b)

Tương tự : 2015+b^2 = ( b + c )(b +a )

 2015 + c^2 = ( c + a )(c + b ) thay vào ta có :

( 2015 + a^2)(2015 + b^2 ) (2015 +c^2) = (a + c )(a+b)(b+c)(b+a)(c+a)(c+b) = [(a+c)(a+b)(b+c) ]^2 là số chính phương 

Câu 1 ) :

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2015}\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{2015}-\frac{1}{z}=\frac{z-2015}{2015z}\)

=> \(\frac{x+y}{xy}=\frac{z-2015}{2015z}\)

=> \(2015z\left(x+y\right)=\left(z-2015\right)xy\)

=> \(2015z\left(2015-z\right)-\left(z-2015\right)xy\) = 0 

=> \(\left(2015-z\right)\left(2015z-xy\right)\)= 0

=> \(\left(2015-z\right)\left(2015\left(2015-x-y\right)-xy\right)=0\)

=> \(\left(2015-z\right)\left(2015^2-2015x-2015y-xy\right)=0\)

=> \(\left(2015-z\right)\left(2015-x\right)\left(2015-y\right)=0\)

=> 2015 - z =  0 hoặc 2015 -x = 0 hoặc 2015 - y = 0 

=> z = 2015 hoặc x= 2015 hoặc y = 2015 

Vậy trong ba số có ít nhất 1 số bằng 2015 

Câu này olm phải chọn câu dưới em vừa làm vừa nghĩ

Từ giả thiết ta có ngay \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)

\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)

\(\Leftrightarrow\left(x+y\right)\left[\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right]=0\)

\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

Suy ra x + y = 0 hoặc y + z = 0 hoặc z + x = 0

Tới đây bạn tự làm nhé :)

     \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\left(x;y;z,x+y+z\ne0\right)\)

\(\Rightarrow\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)

\(\Rightarrow\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

\(\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)

\(\Leftrightarrow\left(xy+yz\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow y\left(x+z\right)\left(x+y+z\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz\right)+xz\left(x+z\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left(xy+y^2+yz+xz\right)=0\)

\(\Leftrightarrow\left(x+z\right)\left[y\left(x+y\right)+z\left(x+y\right)\right]=0\)

\(\Leftrightarrow\left(x+z\right)\left(x+y\right)\left(y+z\right)=0\)

Từ đó \(x=-z\)hoặc \(x=-y\)hoặc \(y=-z\)

-Nếu \(x=-z\Rightarrow z^{2017}+x^{2017}=0\Rightarrow M=\frac{19}{4}+0=\frac{19}{4}\)

Tương tự với các trường hợp còn lại, ta cũng tính được \(M=\frac{19}{4}\)

tự túc

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 )

        \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2015}\)

\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)   (do x+y+z = 2015)

\(\Rightarrow\)\(\frac{xy+yz+xz}{xyz}=\frac{1}{x+y+z}\)

\(\Rightarrow\)\(\left(xy+yz+xz\right)\left(x+y+z\right)=xyz\)

\(\Rightarrow\)\(\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\)

\(\Rightarrow\)\(\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

đến đây tự lm nốt nha

Đặt \(\sqrt{x-2013}=a\left(a>0\right)\)

\(\sqrt{y-2014}=b\left(b>0\right)\)

\(\sqrt{z-2015}=c\left(c>0\right)\)

Có \(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

<=> \(\frac{a-1}{a^2}-\frac{1}{4}+\frac{b-1}{b^2}-\frac{1}{4}+\frac{c-1}{c^2}-\frac{1}{4}=0\)

<=> \(\frac{4a-4-a^2}{4.a^2}+\frac{4b-4-b^2}{4b^2}+\frac{4c-4+c^2}{4c^2}=0\)

<=>\(\frac{-\left(a^2-4a+4\right)}{4a^2}-\frac{b^2-4b+4}{4b^2}-\frac{c^2-4c+4}{4c^2}=0\)

<=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}=0\).

Có \(\frac{\left(a-2\right)^2}{4a^2}\ge0\forall a>0\)

\(\frac{\left(b-2\right)^2}{4b^2}\ge0\forall b>0\)

\(\frac{\left(c-2\right)^2}{4c^2}\ge0\forall c>0\)

=> \(\frac{\left(a-2\right)^2}{4a^2}+\frac{\left(b-2\right)^2}{4b^2}+\frac{\left(c-2\right)^2}{4c^2}\ge0\) với moi a,b,c >0

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a-2=0\\b-2=0\\c-2=0\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}\sqrt{x-2013}=2\\\sqrt{y-2014}=2\\\sqrt{z-2015}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x-2013=4\\y-2014=4\\z-2015=4\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}x=2017\\y=2018\\z=2019\end{matrix}\right.\)(t/m)

Vậy \(\left(x,y,z\right)\in\left\{\left(2017,2018,2019\right)\right\}\)

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