Đáp án:
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`1,`
`|2x-3| - x = |2-x|` `(1)`
$\bullet$ Nếu `2 - x > 0 -> x <2`
`(1)` có dạng :
`↔ |2x-3| -x=2-x`
`↔ |2x-3| = 2`
`↔` \(\left[ \begin{array}{l}2x-3=2\\2x-3=-2\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}x=\dfrac{5}{2} \text{(Không thỏa mãn)}\\x=\dfrac{1}{4} \text{(Thỏa mãn)}\end{array} \right.\)
$\bullet$ Nếu $2 - x \leqslant 0 ↔ x \geqslant 2$
`(1)` có dạng :
`↔ |2x-3| -x=- (2-x)`
`↔ |2x-3| -x=-2+x`
`↔ |2x-3| = -2+x+x`
`↔ |2x-3| = 2x-2`
Kết hợp với điều kiện : $↔ 2x - 2 \geqslant 2 ↔ x \leqslant 2$
$↔ |2x-3| = |2x-2|$
`↔` \(\left[ \begin{array}{l}2x-3=2x-2\\2x-3=-2x+2\end{array} \right.\)
`↔` \(\left[ \begin{array}{l}0=1 \text{(vô lí)}\\x=\dfrac{5}{4} \text{(Không thỏa mãn)} \end{array} \right.\)
Vậy `x=1/4`
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`2,`
`2 |x-3| - |4x-1|=0`
`↔ 2 |x-3| = 0 + |4x-1|`
`↔ 2 |x-3| = |4x-1|`
Trường hợp 1 :
`↔ 2 (x-3) = 4x-1`
`↔2x-6=4x-1`
`↔2x-4x=6-1`
`↔-2x=5`
`↔x=5÷(-2)`
`↔x=(-5)/2`
Vậy `x=(-5)/2`
Trường hợp 2 :
`↔ 2 (x-3) = - (4x-1)`
`↔ 2x - 6 = -4x + 1`
`↔2x+4x=6+1`
`↔6x=7`
`↔x=7÷6`
`↔x=7/6`
Vậy `x ∈ { (-5)/2; 7/6}`
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`3,`
`|x+2| + |2y +3| =0`
Với mọi `x,y` có : \(\left\{ \begin{array}{l}|x+2| \geqslant 0\\|2y+3| \geqslant 0\end{array} \right.\)
$↔ |x+2| + |2y+3| \geqslant 0 ∀ x,y$
Dấu "`=`" xảy ra khi :
`↔` \(\left\{ \begin{array}{l}|x+2|=0\\|2y+3|=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x+2=0\\2y+3=0\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=0-2\\2y=0-3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=-2\\2y=-3\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=-2\\y=-3÷2\end{array} \right.\)
`↔` \(\left\{ \begin{array}{l}x=-2\\y=\dfrac{-3}{2}\end{array} \right.\)
Vậy `(x;y)= (-2; (-3)/2)`
