Thêm điều kiện : x,y,z khác 0 và x+y+z khác 0

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)\(\Rightarrow\) \(\left(\frac{1}{x}+\frac{1}{y}\right)+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\)\(\Leftrightarrow\left(x+y\right)\left(\frac{xz+xy+yz+z^2}{xyz\left(x+y+z\right)}\right)=0\)\(\Leftrightarrow\frac{\left(x+y\right)\left(x+z\right)\left(z+y\right)}{xyz\left(x+y+z\right)}=0\)

Do đó : x + y = 0 hoặc x + z = 0 hoặc y + z = 0

Từ đó thay x,y,z vào từng trường hợp rồi suy ra đpcm

1/x+1/y+1/z=1/xyz

1/x+1/y=1/xyz-1/z

(x+y)(xy+yz+z^2)=0

(x+y)(x+z)(y+z)=0

x+y=0 suy ra x=-y

x+z=o suy ra z=x

z+y=0 suy ra y=-z

voi x=-y suy ra 1/x^2016+1/y^2016+1/z^2016=1/-y^2016+1/y^2016+1/z^2016=1/z^2016 (1)

1/x^2016+y^2016+z^2016=1/-y^2016+y^2016+z^2016 =1/z^2016 (2)

tu 1 va 2 suy ra dpcm

tinh gum minh cai chc chan bai nay dung

Có cách nào dễ hiểu hơn ko

\(\left(\sqrt{x-2016}-2\right)^2+\left(\sqrt{y-2016}-2\right)^2+\left(\sqrt{z-2016}-2\right)=0..\)

=> x=y=z = 2020

có dấu âm đằng trước nữa đúng ko bạn?

cảm ơn bạn nhiều nha 

Ta có :

\(x=\frac{2016^{2017}+1}{2016^{2016}+1}\)

\(\frac{1}{2016}x=\frac{2016^{2017}+1}{2016^{2017}+2016}=\frac{2016^{2017}+2016-2015}{2016^{2017}+2016}\)

\(\Rightarrow\frac{1}{2006}x=1-\frac{2015}{2016^{2017}+2016}\)

Ta lại có :

\(y=\frac{2016^{2016}+1}{2016^{2015}+1}\)

\(\Rightarrow\frac{1}{2016}y=\frac{2016^{2016}+1}{2016^{2016}+2016}=\frac{2016^{2016}+2016-2015}{2016^{2016}+2016}\)

\(\Rightarrow\frac{1}{2016}y=1-\frac{2015}{2016^{2016}+2016}\)

Mà \(\frac{2015}{2016^{2017}+2016}< \frac{2015}{2016^{2016}+2016}\)(so sánh mẫu)

\(\Rightarrow1-\frac{2015}{2016^{2017}+2016}>1-\frac{2015}{2016^{2016}+2016}\)

\(\Rightarrow\frac{1}{2016}x>\frac{1}{2016}y\)

\(\Rightarrow x>y\)

DÀI QUÁ KHÔNG TÍNH ĐƯỢC. CÁI NÀY CÓ MÀ ĐI HỎI THẦN ĐỒNG VỀ MÔN TOÁN ĐI

\(x< \frac{2016^{2017}+1+2015}{2016^{2016}+1+2015}\)

\(\Rightarrow x< \frac{2016^{2017}+2016}{2016^{2016}+2016}\)

\(\Rightarrow x< \frac{2016.\left(2016^{2016}+1\right)}{2016.\left(2016^{2015}+1\right)}\)

\(\Rightarrow x< y\)

. đi bạn

help me

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có:

\(x=\dfrac{2016^{2017}+1}{2016^{2016}+1}< 1\)

\(\Rightarrow x< \dfrac{2016^{2017}+1+2015}{2016^{2016}+1+2015}\Rightarrow x< \dfrac{2016^{2017}+2016}{2016^{2016}+2016}\Rightarrow x< \dfrac{2016\left(2016^{2016}+1\right)}{2016\left(2016^{2015}+1\right)}\Rightarrow x< \dfrac{2016^{2016}+1}{2016^{2015}+1}=y\)

\(\Rightarrow x< y\)