Question

Tìm x,y

|x² - 4 | + | y² - 9 | = 0

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Answer 1

Vì \(\left\{{}\begin{matrix}\left|x^2-4\right|\ge0\\\left|y^2-9\right|\ge0\end{matrix}\right.\)=>|x² - 4 | + | y² - 9 | \(\ge\) 0

Dấu "=" xảy ra khi |x² - 4 | = | y² - 9 | = 0

|x² - 4 |=0 => x²-4=0 => x²=4 => \(x=\pm2\)

| y² - 9 |=0 =>y²-9=0=>y²=9=>\(y=\pm3\)

Vậy \(x=\pm2\) và \(y=\pm3\)

Answer 2

Ta có :

\(\left[{}\begin{matrix}\left|x^2-4\right|\ge0\\\left|y^2-9\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x^2-4\right|+\left|y^2-9\right|\ge0\)

\(x^2-4=0\Rightarrow x=\pm2\)

\(y^2-9=0\Rightarrow y=\pm3\)

Vậy......................