+)Xét x+y+z khác 0

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

\(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-3}=\frac{x+y+z}{y+z+1+x+z+2+x+y-3}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

=>x+y+z=1/2

\(\Rightarrow\hept{\begin{cases}2x=y+z+1\\2y=x+z+2\\2z=x+y-3\end{cases}\Rightarrow\hept{\begin{cases}3x=x+y+z+1\\3y=x+y+z+2\\3z=x+y+z-3\end{cases}\Rightarrow}\hept{\begin{cases}3x=\frac{1}{2}+1\\3y=\frac{1}{2}+2\\3z=\frac{1}{2}-3\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=\frac{-5}{6}\end{cases}}}\)

+)Xét x+y+z=0

=>x/y+z+1=y/x+z+2=z/x+y-3=0

=>x=y=z=0

\(\frac{x^2}{2y}+\frac{y^2}{2x}+\frac{y^2}{2z}+\frac{z^2}{2y}+\frac{z^2}{2x}+\frac{x^2}{2z}\ge\frac{\left(2x+2y+2z\right)^2}{4\left(x+y+z\right)}=x+y+z\)

bạn đưa về 1 ẩn rồi giải nhen :

a) \(\frac{x}{y}=\frac{2}{3}\Rightarrow y=\frac{3x}{2}\)

Ta có : \(x.y=54\Leftrightarrow x.\frac{3x}{2}=54\)

\(\Rightarrow3x^2=108\)

\(\Rightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

Giải:

Ta có:

\(\frac{x}{2}=\frac{y}{3};\frac{y}{2}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6};\frac{y}{6}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)

Theo tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)

+) \(\frac{x}{4}=2\Rightarrow x=8\)

+) \(\frac{y}{6}=2\Rightarrow y=12\)

+) \(\frac{z}{15}=2\Rightarrow z=30\)

Vậy x = 8

       y = 12

       z = 30

\(\frac{x}{2}=\frac{y}{3};\frac{y}{2}=\frac{z}{5}\) và x + y + z =50

\(\frac{x}{4}=\frac{y}{6};\frac{y}{6}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{x}{4}+\frac{y}{6}+\frac{z}{15}=\frac{50}{25}=2\)

=> x = 2.4 = 8

=> y = 2.6 = 12

=> z = 2.15 = 30

Vậy x = 8;y = 12;z = 30. 

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\) và \(\frac{y}{2}=\frac{z}{5}\Rightarrow\frac{y}{6}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{15}\)

Áp dụng tc dãy tỉ = nhau 

\(\frac{x}{4}=\frac{y}{6}=\frac{z}{15}=\frac{x+y+z}{4+6+15}=\frac{50}{25}=2\)

bạn ơi đề thiếu

Theo tính chất của dãy tỉ số bằng nhau, ta có

\(\frac{y+z+1}{x}=\frac{x+y+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}=\frac{y+z+1+x+y+2+x+y-3+1}{x+y+z+x+y+z}\)

=\(\frac{\left(x+y+z\right)+\left(x+y+y1+2-3\right)}{\left(x+y+z\right)+\left(x+y+z\right)}=\frac{\left(x+y+z\right)+\left(x+y+y+1\right)}{\left(x+y+z\right)+\left(x+y+z\right)}\)

=>x+y+y+1=x+y+z

=>y+1=z

Vậy đáp số cần tìm là x,y,z khác 0

x tùy ý

y tùy ý

z=y+1

Đề là gì ????

Tìm x,y,z

  Bài 1:  \(x\).(\(x-y\)) = \(\dfrac{3}{10}\) và y(\(x-y\)) = - \(\dfrac{3}{50}\)

    \(x\)(\(x\) - y) - y(\(x\) - y) = \(\dfrac{3}{10}\) - ( - \(\dfrac{3}{50}\))

     (\(x-y\)).(\(x-y\)) = \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)

        (\(x-y\))² = \(\dfrac{15}{50}\) + \(\dfrac{3}{50}\)

        (\(x\) - y)² = \(\dfrac{9}{25}\) = (\(\dfrac{3}{5}\))²

        \(\left[{}\begin{matrix}x-y=-\dfrac{3}{5}\\x-y=\dfrac{3}{5}\end{matrix}\right.\) 

TH1 \(x-y=-\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\left(-\dfrac{3}{5}\right)=\dfrac{3}{10}\\y.\left(-\dfrac{3}{5}\right)=-\dfrac{3}{50}\end{matrix}\right.\) 

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\left(-\dfrac{3}{5}\right)=\dfrac{-1}{2}\\y=-\dfrac{3}{50}:\left(-\dfrac{3}{5}\right)=\dfrac{1}{10}\end{matrix}\right.\) 

TH2: \(x-y=\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\dfrac{3}{5}=\dfrac{3}{10}\\y.\dfrac{3}{5}=-\dfrac{3}{50}\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\dfrac{3}{5}=\dfrac{1}{2}\\y=-\dfrac{3}{50}:\dfrac{3}{5}=-\dfrac{1}{10}\end{matrix}\right.\)  

    Vậy (\(x;y\)  ) = (- \(\dfrac{1}{2}\); \(\dfrac{1}{10}\)); (\(\dfrac{1}{2}\); - \(\dfrac{1}{10}\))