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

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

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

g,

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

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

Áp dụng tính chất của dãy tỉ số bằng nhau

\(\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}=\dfrac{15x-10y+6z-15x+10y-6z}{25+9+4}=0\)\(\Rightarrow3x-2y=2z-5x=5y-3z=0\)

* 3x - 2y = 0 \(\Rightarrow3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\)

* 2z - 5x = 0 \(\Rightarrow2z=5x\Rightarrow\dfrac{x}{2}=\dfrac{z}{5}\)

Áp dụng tính chất của dãy tỉ số bằng nhau

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

\(\cdot\dfrac{x}{2}=5\Rightarrow x=10\)

\(\cdot\dfrac{y}{3}=5\Rightarrow y=15\)

\(\cdot\dfrac{z}{5}=5\Rightarrow z=25\)

câu h thiếu điều kiện rồi bạn ơi

Giải:
Ta có: \(\dfrac{3x-2y}{5}=\dfrac{5y-3z}{2}=\dfrac{2z-5x}{2}\)

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

Áp dụng tính chất dãy tỉ số bằng nhau có:

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

\(\Rightarrow\left\{{}\begin{matrix}15x-10y=0\\10y-6z=0\\6z-15x=0\end{matrix}\right.\Rightarrow15x=10y=6z\)

\(\Rightarrow\dfrac{15x}{30}=\dfrac{10y}{30}=\dfrac{6z}{30}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=10\\y=15\\z=25\end{matrix}\right.\)

Vậy...

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

\(\Rightarrow\dfrac{5\left(3x-2y\right)}{25}=\dfrac{2\left(5y-3z\right)}{4}=\dfrac{3\left(2z-5x\right)}{6}\)

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

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

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

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

\(=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\\5y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{5}\\2z=5x\Rightarrow\dfrac{z}{5}=\dfrac{x}{2}\end{matrix}\right.\)

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

\(\Rightarrow\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}\)

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

\(\dfrac{10x}{20}=\dfrac{3y}{9}=\dfrac{2z}{10}=\dfrac{10x-3y-2z}{20-9-10}=\dfrac{5}{1}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.5=25\end{matrix}\right.\)

a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)

Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)

\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)

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

\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)

\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)

\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)

Xin lỗi mình chỉ làm được câu a)

 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

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

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

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

\(\dfrac{3x-2y}{5}=\dfrac{2z-5x}{3}=\dfrac{5y-3z}{2}=\dfrac{15x-10y}{25}=\dfrac{6z-15x}{9}=\dfrac{10y-6z}{4}\)

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

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

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

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

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{2x}{6}=\dfrac{3y}{9}=\dfrac{5z}{25}\)\(=\dfrac{2x+3y-5z}{6+9-25}=\dfrac{-60}{-10}=6\)

⇒\(\dfrac{x}{2}=6\)⇒\(x=12\)

⇒\(\dfrac{y}{3}=6\)⇒\(y=18\)

⇒\(\dfrac{z}{5}=6\)⇒\(z=30\)

Vậy \(x=12;y=18;z=30\)

biết??????

b tham khảo nhé