Đề không đầy đủ đề tìm $x,y$ cụ thể. Bạn xem lại.

y x 10 + y + 10 = 200

y x ( 10 + 10 )   = 200

y x 20               = 200

y                      = 200 : 20

y                      = 10

y x 10 + y x 10 = 200

y x ( 10 + 10 ) = 200

y x 20 = 200

y = 200 : 20

y = 10 

y x 10 + y x 10 = 200

y x ( 10 + 10 ) = 200

y x 20 = 200

y = 200 : 20

y = 10

Mk nhanh nhất đó

**** mk nha 

y x 10 +  x 10 = 200

 y x ( 10 + 10 ) =200

 y x 20 =200

 y = 200  : 20

 y = 10

=> y x ( 4 + 6 ) = 20

=> y x 10 =20

=> y = 2

y x 4 + y x 6 = 200 : 10

y x ( 4 + 6 ) = 20

y x 10 = 20

y = 20 : 10

y = 2

a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)

               \(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)

\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)

b)Ta có:  \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

               \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)

Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)

2070 ; 2970 ; 2475 nha bạn

tk mk nha

cả x lẫn y đều là 0 nha bạn

Ta có \(\frac{x}{5}=\frac{y}{9}=\frac{z}{3}\Rightarrow\frac{x}{5}=\frac{y}{9}=\frac{-z}{-3}\)

\(\Rightarrow\frac{x}{5}=\frac{x+y-z}{5+9-3}=\frac{x+\left(y-z\right)}{11}=\frac{x+18}{11}\)

\(\Rightarrow\frac{x}{5}=\frac{x+18}{11}\Rightarrow11x=5x+90\Rightarrow x=15\)

\(\Rightarrow\frac{y}{9}=\frac{15}{5}\Rightarrow y=27\); \(\frac{z}{3}=3\Rightarrow z=9\)

Vậy \(x=15;y=27;z=9\)

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