Question

[Nguyễn Văn Thiện]

1 tìm x,y biết 

a, /x-y/+/y+\(\frac{9}{25}\)/ =0

/     / là trị tiệt đối

Answer 1

[Arima Kousei]

Đặt \(A=\left|x-y\right|+\left|y+\frac{9}{25}\right|\)

\(\hept{\begin{cases}\left|x-y\right|\ge0\forall x;y\\\left|y+\frac{9}{25}\right|\ge0\forall x;y\end{cases}\Rightarrow A=\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\forall x;y}\)

Dấu \("="\)

\(\Leftrightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|y+\frac{9}{25}\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\y=-\frac{9}{25}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\y=-\frac{9}{25}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-\frac{9}{25}\\y=-\frac{9}{25}\end{cases}}}\)

Vậy \(\hept{\begin{cases}x=-\frac{9}{25}\\y=-\frac{9}{25}\end{cases}}\)

Answer 2

[kudo shinichi]

\(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)

Ta có: \(\hept{\begin{cases}\left|x-y\right|=0\forall x;y\\\left|y+\frac{9}{25}\right|=0\forall y\end{cases}\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\forall x;y}\)

Mà \(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x-y\right|=0\\\left|y+\frac{9}{25}\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-y=0\\y+\frac{9}{25}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\y=-\frac{9}{25}\end{cases}\Rightarrow}x=y=-\frac{9}{25}}\)

Vậy \(x=y=-\frac{9}{25}\)