a,/2x-1/+5/y+2 =0 B, /2x-4/+/x+y-4/=0 C, /x-1/+/x ×y +5/=0 C, /x-y+2/ +/4-y/=0

Đáp án:

$\\$

`a,`

`|2x – 1| + 5 |y + 2| = 0`

Do : \(\left\{ \begin{array}{l}|2x-1|\geqslant 0∀x\\|y+2|\geqslant 0∀y\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}|2x-1|\geqslant 0∀x\\5|y+2|\geqslant 0∀y\end{array} \right.\)

$↔ |2x-1| +  5 |y + 2| \geqslant 0 ∀ x,y$

Dấu “`=`” xảy ra khi :

`↔` \(\left\{ \begin{array}{l}2x-1=0\\y+2=0\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}2x=1\\y=-2\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=\dfrac{1}{2}\\y=-2\end{array} \right.\)

Vậy `(x;y) = (1/2; -2)`

$\\$

`b,`

`|2x-4| + |x + y – 4| = 0`

Do : \(\left\{ \begin{array}{l}|2x-4|\geqslant 0∀x\\|x+y-4|\geqslant 0∀y\end{array} \right.\)

$↔ |2x-4| + |x  +y-4| \geqslant 0 ∀ x,y$

Dấu “`=`” xảy ra khi :

`↔` \(\left\{ \begin{array}{l}2x-4=0\\x+y-4=0\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}2x=4\\x+y=4\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=2\\2+y=4\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=2\\y=2\end{array} \right.\)

Vậy `(x;y)=(2;2)`

$\\$

`c,`

`|x-1| + |xy  + 5| = 0`

Do : \(\left\{ \begin{array}{l}|x-1|\geqslant 0∀x\\|xy+5|\geqslant 0∀y\end{array} \right.\)

$↔ |x-1| + |xy + 5| \geqslant 0∀x,y$

Dấu “`=`” xảy ra khi :

`↔` \(\left\{ \begin{array}{l}x-1=0\\xy+5=0\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=1\\1.y=-5\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=1\\y=-5\end{array} \right.\)

Vậy `(x;y) = (1;-5)`

$\\$

`d,`

`|x-y+2| + |4-y| = 0`

Do : \(\left\{ \begin{array}{l}|x-y+2|\geqslant 0∀x\\|4-y|\geqslant 0∀y\end{array} \right.\)

$↔ |x-y+2| + |4-y| \geqslant 0 ∀ x,y$

Dấu “`=`” xảy ra khi :

`↔` \(\left\{ \begin{array}{l}x-y+2=0\\4-y=0\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x-4=-2\\y=4\end{array} \right.\)

`↔` \(\left\{ \begin{array}{l}x=2\\y=4\end{array} \right.\)

Vậy `(x;y) = (2;4)`

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