9 tìm x,y thuộc Z a, /x-8/+/Y+2/ b, xy= x+y c, xy= x-y 08/11/2021 Bởi Serenity 9 tìm x,y thuộc Z a, /x-8/+/Y+2/ b, xy= x+y c, xy= x-y
Đáp án: `↓↓` Giải thích các bước giải: `a) | x – 8 | + | y + 2 |` Ta có: `| x – 8 | >= 0 ∀x` `| y + 2 | >= 0 ∀y` `=> | x – 8 | + | y + 2 | >=0 ∀x,y` Dấu “=” xảy ra `<=>` $\left\{\begin{matrix}|x-8| = 0& \\|y+2| = 0& \end{matrix}\right.$⇒ $\left\{\begin{matrix}x-8 = 0& \\y+2 = 0& \end{matrix}\right.$⇒ $\left\{\begin{matrix}x=8 & \\y=-2 & \end{matrix}\right.$ `b) xy = x+y` `=> xy – x – y = 0 ` `=> x ( y – 1 ) -1 ( y – 1 ) – 1 = 0` `=> ( x – 1 ) ( y – 1 ) = 1` Ta có bảng: $\begin{array}{|c|c|}\hline x-1&1&-1\\\hline y-1&1&-1\\\hline x&2&0 \\\hline y&2&0\\\hline \end{array}$ `c) xy = x – y` `=> xy – x + y = 0` `=> x ( y – 1 ) + 1 ( y – 1 ) + 1 = 0` `=> ( x + 1 ) ( y – 1 ) = -1` Ta có bảng: $\begin{array}{|c|c|}\hline x+1&1&-1\\\hline y-1&-1&1\\\hline x&0&-2 \\\hline y&0&2\\\hline\end{array}$ Bình luận
Đáp án:
`↓↓`
Giải thích các bước giải:
`a) | x – 8 | + | y + 2 |`
Ta có: `| x – 8 | >= 0 ∀x`
`| y + 2 | >= 0 ∀y`
`=> | x – 8 | + | y + 2 | >=0 ∀x,y`
Dấu “=” xảy ra `<=>` $\left\{\begin{matrix}|x-8| = 0& \\|y+2| = 0& \end{matrix}\right.$
⇒ $\left\{\begin{matrix}x-8 = 0& \\y+2 = 0& \end{matrix}\right.$
⇒ $\left\{\begin{matrix}x=8 & \\y=-2 & \end{matrix}\right.$
`b) xy = x+y`
`=> xy – x – y = 0 `
`=> x ( y – 1 ) -1 ( y – 1 ) – 1 = 0`
`=> ( x – 1 ) ( y – 1 ) = 1`
Ta có bảng:
$\begin{array}{|c|c|}\hline x-1&1&-1\\\hline y-1&1&-1\\\hline x&2&0 \\\hline y&2&0\\\hline \end{array}$
`c) xy = x – y`
`=> xy – x + y = 0`
`=> x ( y – 1 ) + 1 ( y – 1 ) + 1 = 0`
`=> ( x + 1 ) ( y – 1 ) = -1`
Ta có bảng:
$\begin{array}{|c|c|}\hline x+1&1&-1\\\hline y-1&-1&1\\\hline x&0&-2 \\\hline y&0&2\\\hline\end{array}$