\(\dfrac{3x-2y}{5}\)=\(\dfrac{2z-5x}{3}\)=\(\dfrac{5y-3z}{2}\)

⇒\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{15x-10y}{25}\)=\(\dfrac{6z-15x}{9}\)=\(\dfrac{10y-6z}{4}\)=\(\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}\)=0

⇒3x-2y=2z-5x=5y-3z=0

* 3x-2y=0⇒3x=2y⇒\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\) 

* 2z-5x=0⇒2z=5x⇒\(\dfrac{z}{5}\)=\(\dfrac{x}{2}\) 

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)=\(\dfrac{x+y+z}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5

\(\dfrac{x}{2}\)=-5⇒x=-10

\(\dfrac{y}{3}\)=-5⇒y=-15

\(\dfrac{z}{5}\)=-5⇒z=-25

Vậy x=-10;y=-15;z=-25

Từ đẳng thức : \(\frac{3x-2y}{5}=\frac{2z-5x}{3}=\frac{5y-3z}{2}\)

=> \(\frac{15x-10y}{5^2}=\frac{6z-15x}{3^2}=\frac{10y-6z}{2^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có : 

\(\frac{15x-10y}{5^2}=\frac{6z-15x}{3^2}=\frac{10y-6z}{2^2}=\frac{15x-10y+6z-15x+10y-6z}{5^2+3^2+2^2}=0\)

=> \(\hept{\begin{cases}15x=10y\\6z=15x\\10y=6z\end{cases}\Rightarrow\hept{\begin{cases}3x=2y\\2z=5x\\5y=3z\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{z}{5}=\frac{x}{2}\\\frac{y}{3}=\frac{z}{5}\end{cases}\Rightarrow}\frac{x}{2}=\frac{y}{3}=\frac{z}{5}}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Khi đó : x² + 176 = yz 

<=> (2k)² - 15k² = -176

=> k²(4 - 15) = -176

=> k² = 16

=> k² = 4²

=> k = \(\pm\)4

Nếu k = 4 

=> \(\hept{\begin{cases}x=8\\y=12\\z=20\end{cases}}\)

Nếu k = - 4

=> \(\hept{\begin{cases}x=-8\\y=-12\\z=-20\end{cases}}\)

 đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)

\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)

BBDT AM-GM 

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)

theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)

vì \(x^2+y^2\ge2xy\)

\(y^2+z^2\ge2yz\)

\(x^2+z^2\ge2xz\)

\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)

\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)

\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)

dấu"=" xảy ra<=>x=y=z=1/3

=)))))))))))))))))))))))))))))))))))))))))))))))))))))))

Áp dụng tính chất dãy tỉ số bằng nhau:\(\dfrac{\left(5z-3y\right)+\left(3x-2z\right)+\left(2y-5x\right)}{2+5+3}\)

=\(\dfrac{\left(3x-5x\right)+\left(-3y+2y\right)+\left(5z-2z\right)}{2+5+3}\)

=\(\dfrac{-2x-y+3z}{2+5+3}\)(???!!!!)

=\(\dfrac{-2x}{2}=\dfrac{-y}{5}=\dfrac{3z}{3}\)

=\(\dfrac{2}{-2x}=\dfrac{5}{-y}=\dfrac{3}{3z}\)

tớ xin chịu trận vì ko chứng minh được :(((

nó lại ra như thế này