a: x/7=9/y

nên xy=63

mà x>y

nên \(\left(x,y\right)\in\left\{\left(63;1\right);\left(21;3\right);\left(9;7\right);\left(-1;-63\right);\left(-3;-21\right);\left(-7;-9\right)\right\}\)

b: -2/x=y/6

nên xy=-12

mà x<0<y

nên \(\left(x,y\right)\in\left\{\left(-12;1\right);\left(-6;2\right);\left(-4;3\right);\left(-3;4\right);\left(-2;6\right);\left(-1;12\right)\right\}\)

a) thay \(x-y=\frac{3}{10}\)vào \(y\left(x-y\right)=\frac{-3}{50}\)ta có\(\frac{3}{10}y=\frac{-3}{50}\)=>\(y=\frac{-3}{50}:\frac{3}{10}=\frac{-1}{5}\)=>\(x-y=\frac{3}{10}\Rightarrow x=\frac{3}{10}+\frac{-1}{5}=\frac{1}{10}\)

hôm sau mik giải tip cho

Ta có: \(\left(x-2\right)^{2012}\ge0\forall x\)

\(\left|y-9\right|^{2014}\ge0\forall y\)

\(\Rightarrow\left(x-2\right)^{2012}+\left|y-9\right|^{2014}=0\Leftrightarrow\left(x-2\right)^{2012}=\left|y-9\right|^{2014}=0\)

\(\Rightarrow x-2=y-9=0\)

\(\Rightarrow x=2\)và \(y=9\)

Vậy x = 2; y = 9

(x-5)(y+1)=6

Ta có bảng

Vậy các cặp (x;y) là (6;5) ; (7;2) ; (8;1) ; (11;0)

\(\left(x-5\right)\left(x^2.9\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-5=0\\x^2.9=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x^2=0\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=5\\x=0\end{cases}}\)