Đáp án:

`a)|2x-1|+5|y+2|=0`

Vì \(\begin{cases}|2x-1| \ge 0\\5|y+2 \ge 0\\\end{cases}\)

`=>|2x-1|+5|y+2|>=0`

Dấu "=" xảy ra khi:\(\begin{cases}2x-1=0\\5(y+2)=0\\\end{cases}\)

`<=>` \(\begin{cases}2x=1\\y+2=0\\\end{cases}\)

`<=>` \(\begin{cases}x=\dfrac12\y=-2\\\end{cases}\)

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

Vì \(\begin{cases}|2x-4| \ge 0\\|x+y-4| \ge 0\\\end{cases}\)

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

Dấu "=" xảy ra khi:\(\begin{cases}2x-4=0\\x+y-4=0\\\end{cases}\)

`<=>` \(\begin{cases}2x=4\\y=4-x\\\end{cases}\)

`<=>` \(\begin{cases}x=2\\y=2\\\end{cases}\)

`c)|x-1|+|xy+5|=0`

Vì \(\begin{cases}|x-1| \ge 0\\|xy+5| \ge 0\\\end{cases}\)

`=>|x-1|+|xy+5|>=0`

Dấu "=" xảy ra khi:\(\begin{cases}x-1=0\\xy+5=0\\\end{cases}\)

`<=>` \(\begin{cases}x=1\\y=-5\\\end{cases}\)

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

Vì \(\begin{cases}|x-y+2| \ge 0\\|4-y| \ge 0\\\end{cases}\)

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

Dấu "=" xảy ra khi:\(\begin{cases}x-y+2=0\\4-y=0\\\end{cases}\)

`<=>` \(\begin{cases}y=4\\x=y-2=2\\\end{cases}\)

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)`