Question

Tìm x và y biết:

\(\left|x+\frac{2}{5}\right|+\left|2y-\frac{1}{3}\right|=0\)

Answer 1

Ta có | x + 2/5 | ≥ 0 ∀ x

         | 2y - 1/3 | ≥ 0 ∀ y

=> | x + 2/5 | + | 2y - 1/3 | ≥ 0 ∀ x, y

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+\frac{2}{5}=0\\2y-\frac{1}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{5}\\y=\frac{1}{6}\end{cases}}\)

Vậy x = -2/5 ; y = 1/6

Answer 2

\(\left|x+\frac{2}{5}\right|+\left|2y-\frac{1}{3}\right|=0\)

\(\orbr{\begin{cases}\left|x+\frac{2}{5}\right|=0\\\left|2y-\frac{1}{3}\right|=0\end{cases}}\)

\(\orbr{\begin{cases}x=0-\frac{2}{5}\\2y=0+\frac{1}{3}\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{2}{5}\\2y=\frac{1}{3}\end{cases}}\)

\(x=\frac{1}{3}:2\)

\(x=\frac{2}{3}\)

vậy \(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{2}{3}\end{cases}}\)

Answer 3

nhầm nha

\(\left|x+\frac{2}{5}\right|+\left|2y-\frac{1}{3}\right|=0\)

\(\orbr{\begin{cases}\left|x+\frac{2}{5}\right|=0\\\left|2y-\frac{1}{3}\right|=0\end{cases}}\)

\(\orbr{\begin{cases}x=0-\frac{2}{5}\\2y=0+\frac{1}{3}\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{2}{5}\\2y=\frac{1}{3}\end{cases}}\)

\(y=\frac{1}{3}:2\)

\(y=\frac{1}{6}\)

vậy \(\orbr{\begin{cases}x=-\frac{2}{5}\\y=\frac{1}{6}\end{cases}}\)

vậy