Question

[Đoàn Thị Mai Phương]

tìm x,y biết

a) \(|y-1|+|5-x|\)\(\le\)0

b) \(\frac{|y-6|}{x+2}\)> 0

c)\(\frac{x^2-1}{x^2}\)>0

Answer 1

[kudo shinichi]

a) Ta có: \(\hept{\begin{cases}\left|y-1\right|\ge0\forall y\\\left|5-x\right|\ge0\forall x\end{cases}\Rightarrow\left|y-1\right|+\left|5-x\right|\ge0\forall}x;y\)

Mà \(\left|y-1\right|+\left|5-x\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|y-1\right|=0\\\left|5-x\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}y-1=0\\5-x=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\x=5\end{cases}}}\)

Vậy \(\hept{\begin{cases}y=1\\x=5\end{cases}}\)

b)  Ta có: \(\left|y-6\right|\ge0\forall y\)

\(\Rightarrow\left|y-6\right|>0\Leftrightarrow y\ne6\)

\(\Rightarrow\)Để \(\frac{\left|y-6\right|}{x+2}>0\)thì \(\hept{\begin{cases}y\ne6\\x+2>0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y\ne6\\x>-2\end{cases}}\)

Vậy \(\hept{\begin{cases}y\ne6\\x>-2\end{cases}}\)

c) Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow x^2>0\Leftrightarrow x\ne0\)

Để \(\frac{x^2-1}{x^2}>0\Leftrightarrow\hept{\begin{cases}x^2-1>0\\x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x\ne0\end{cases}\Leftrightarrow}x>1}\)

Vậy \(x>1\)

Tham khảo nhé~