Đáp án:
Bài `4.31`
`a)`
`|2x-5|=4`
`<=>`\(\left[ \begin{array}{l}2x-5=4\\2x-5=-4\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=4+5\\2x=-4+5\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=4+5\\2x=-4+5\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{9}{2}\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `S={1/2;9/2}`
`b)`
`3/4 -|2x+1|=7/8`
`<=>3/4-7/8=|2x+1|`
`<=>6/8-7/8=|2x+1|`
`<=>-1/8=|2x+1|` (vô lí) [`∀x` ta có `|2x+1|>=0` mà `-1/8<0`]
`=>` Phương trình vô nghiệm.
Vậy `S=∅`
Bài `4.32`
`a)`
`6,5-9/4:|x+1/3|=2`
`<=>13/2-9/4:|x+1/3|=2`
`<=>13/2 -2=9/4 :|x+1/3|`
`<=>9/2=9/4:|x+1/3|`
`<=>|x+1/3|=9/4:9/2`
`<=>|x+1/3|=9/4 .2/9`
`<=>|x+1/3|=1/2`
`<=>`\(\left[ \begin{array}{l}x+\dfrac{1}{3}=\dfrac{1}{2}\\x+\dfrac{1}{3}=-\dfrac{1}{2}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{2}-\dfrac{1}{3}\\x=-\dfrac{1}{2}-\dfrac{1}{3}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{array} \right.\)
Vậy `S={1/6;-5/6}`
`b)`
`11/4+3/2:|4x-1/5|=7/2`
`<=>3/2 :|4x-1/5|=7/2-11/4`
`<=>3/2 : |4x-1/5|=3/4`
`<=>|4x-1/5|=3/2:3/4`
`<=>|4x-1/5|=3/2 . 4/3`
`<=>|4x-1/5|=2`
`<=>`\(\left[ \begin{array}{l}4x-\dfrac{1}{5}=2\\4x-\dfrac{1}{5}=-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}4x=2+\dfrac{1}{5}\\4x=-2+\dfrac{1}{5}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}4x=\dfrac{11}{5}\\4x=-\dfrac{9}{5}\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{11}{20}\\4x=-\dfrac{9}{20}\end{array} \right.\)
Vậy `S={11/20;-9/20}`
Bài `4.33`
`a)`
`|2x-3|-|3x+2|=0`
`<=>|2x-3|=|3x+2|`
`<=>`\(\left[ \begin{array}{l}2x-3=3x+2\\2x-3=-(3x+2)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x-3=3x+2\\2x-3=-3x-2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}3x-2x=-3-2\\2x+3x=-2+3\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-5\\5x=1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-5\\x=\dfrac{1}{5}\end{array} \right.\)
Vậy `S={-5;1/5}`
`b)`
`|7x+1|-|5x+6|=0`
`<=>|7x+1|=|5x+6|`
`<=>`\(\left[ \begin{array}{l}7x+1=5x+6\\7x+1=-(5x+6)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}7x+1=5x+6\\7x+1=-5x-6\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}7x-5x=6-1\\7x+5x=-6-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=5\\12x=-7\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=-\dfrac{7}{12}\end{array} \right.\)
Vậy `S={5/2;-7/12}`
