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 03/07/2021 Bởi Brielle 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)` Bình luận
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:\(\begin{cases}x-y+2=0\\4-y=0\\\end{cases}\) `<=>` \(\begin{cases}y=4\\x=y-2=2\\\end{cases}\) Bình luận
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)`
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:\(\begin{cases}x-y+2=0\\4-y=0\\\end{cases}\)
`<=>` \(\begin{cases}y=4\\x=y-2=2\\\end{cases}\)