Question

tìm (x,y) nguyên biết :

\(\frac{x-1}{3}+\frac{1}{y}=\frac{-1}{6}\)

\(\frac{x-1}{3}+\frac{1}{y-2}=\frac{-3}{4}\)

nhanh nhất sẽ được tick nhé

Answer 1

\(\frac{x-1}{3}+\frac{1}{y}=\frac{-1}{6}\) 

\(\frac{\left(x-1\right)y}{3y}+\frac{3}{3y}=\frac{-1}{6}\) 

\(\frac{\left(x-1\right)y+3}{3y}=\frac{-1}{6}\) 

\(\frac{\left(x-1\right)y}{y}=\frac{\left(-1\right)-3}{6:3}\) 

\(x-1=-2\)

\(x=\left(-2\right)+1\)

\(x=-1\)

Answer 2

\(\frac{x-1}{3}+\frac{1}{y}=\frac{-1}{6}\)

 \(\frac{\left(x-1\right)y}{3y}+\frac{3}{3y}=\frac{-1}{6}\)

\(\frac{\left(x-1\right)y+3}{3y}=\frac{-1}{6}\) 

\(x-1=\frac{\left(-1\right)-3}{6:3}\)

\(x-1=-2\)

\(x=\left(-2\right)+1\) 

\(x=-1\)

Answer 3

bạn ơi câu thứ hai chưa làm

Answer 4

câu thứ 2 đây nha phanthilan:

\(\frac{x-1}{3}+\frac{1}{y-2}=\frac{-3}{4}\)

\(\frac{x}{3}-\frac{1}{3}+\frac{1}{y-2}=\frac{-3}{4}\)

\(\frac{x}{3}+\frac{1}{y-2}-\frac{1}{3}=\frac{-3}{4}\)

\(\frac{x}{3}+\frac{1}{y-2}=\frac{-3}{4}+\frac{1}{3}\)

\(\frac{x}{3}+\frac{1}{y-2}=\frac{-5}{12}\)

\(\frac{x\left(y-2\right)}{3\left(y-2\right)}+\frac{3}{3\left(y-2\right)}=\frac{-5}{12}\)

\(\frac{x\left(y-2\right)+3}{3\left(y-2\right)}=-\frac{5}{12}\)\(\)

còn lại tự bn suy nghĩ nha,mình vội lắm