\(\dfrac{x-y}{2x+y}=\dfrac{1}{3}\) \(\Rightarrow3\left(x-y\right)=2x+y\)

\(\Rightarrow3x-3y=2x+y\)

\(\Rightarrow x=4y\)

\(\Rightarrow T=\dfrac{x^2}{x^2+y^2}=\dfrac{\left(4y\right)^2}{\left(4y\right)^2+y^2}=\dfrac{16y^2}{16y^2+y^2}=\dfrac{16y^2}{17y^2}=\dfrac{16}{17}\)

(3\(x\) - 5)²⁰⁰⁶ + (y - 1)²⁰⁰⁸ + (\(x\) - 2z)²¹⁰⁰ = 0

Vì (3\(x\) - 5)²⁰⁰⁶ ≥ 0; (y - 1)²⁰⁰⁸ ≥ 0;  (\(x\) - 2z)²¹⁰⁰ ≥ 0 ∀ \(x;y;z\)

Vậy (3\(x-5\))²⁰⁰⁶ + (y - 1)²⁰⁰⁸ + (\(x\) - 2z)²¹⁰⁰ = 0 

⇔ \(\left\{{}\begin{matrix}3x-5=0\\y-1=0\\x-2z=0\end{matrix}\right.\)

⇔ \(\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=1\\z=\dfrac{x}{2}=\dfrac{5}{6}\end{matrix}\right.\)

Vậy...

\(\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=1,5y\)

\(2x+5y=32\)

\(2\times1,5y+5y=32\)

\(3y+5y=32\)

\(\left(3+5\right)y=32\)

\(8y=32\)

\(y=32:8=4\)

⇒ \(x=4\times\dfrac{3}{2}=6\)

Vậy \(x=6\) ; \(y=4\)

\(\dfrac{x}{3}\) = \(\dfrac{y}{4}\); \(\dfrac{y}{5}\) = \(\dfrac{z}{7}\); 2\(x\) + y - z = 372

\(x\) = \(\dfrac{y}{4}\).3; z = \(\dfrac{y}{5}\).7 

Thay \(x\) = \(\dfrac{y}{4}\).3 và z = \(\dfrac{y}{5}\).7 vào biểu thức 2\(x\) + y - z = 372 ta có:

  2.\(\dfrac{y}{4}\).3 + y  - \(\dfrac{y}{5}\).7 = 372

y.( 2.\(\dfrac{1}{4}\).3 + 1 - \(\dfrac{7}{5}\)) = 372

y.\(\dfrac{11}{10}\)    = 372

y            = 372 : \(\dfrac{11}{10}\)

y            = \(\dfrac{3720}{11}\)

\(x\) = \(\dfrac{\dfrac{3720}{11}}{4}\).3 = \(\dfrac{2790}{11}\)

z = \(\dfrac{\dfrac{3720}{11}}{5}\).7 = \(\dfrac{5208}{11}\)

Vậy (\(x;y;z\)) = (\(\dfrac{2790}{11}\); \(\dfrac{3720}{11}\); \(\dfrac{5208}{11}\))

theo đề bài ta có: x/y=4/7 => x= \(\dfrac{4}{7}y\)

Thay vào biểu thức x.y=112

=>\(\dfrac{4}{7}y^2\)=112

<=>\(y^2\)=112:\(\dfrac{4}{7}\)

<=>\(y^2\)=196

<=>\(\left[{}\begin{matrix}y=14\\y=-14\end{matrix}\right.\)

Với y=14 => x=\(\dfrac{4}{7}.14\)=\(8\)

VỚi y=-14 => x=\(\dfrac{4}{7}.\left(-14\right)\)=-8