Question

[Hà Ngọc Linh]

Tìm x và y
a, \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)
b,\(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\)\(\le0\)

Answer 1

[Hắc Hường]

Giải:

Vì:

\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|\ge0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi:

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\le0\)

Vì:

\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|\ge0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

Dấu "=" xảy ra, khi và chỉ khi:

\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|=0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{1}{5}y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{1}{9}\\\dfrac{1}{5}y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{5}{2}\end{matrix}\right.\)

Vậy ...