Bài `1`: Tính:
`a,|-3,7|=3,7`
`b,|(-2)/9|=2/9`
`c,|2 1/3|=2 1/3 = (2.3+1)/3 = 7/3`
`d)|0|=0` ( giá trị tuyệt đối của số tự nhiên vẫn bằng chính số đó , `|a|=a` )
Bài `2:` Tìm `x` biết :
`a,|x+7/8|=5/9`
`=>` \(\left[ \begin{array}{l}x+\dfrac{7}{8}=\dfrac{5}{9}\\x+\dfrac{7}{8}=-\dfrac{5}{6}\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{5}{9}-\dfrac{7}{8}\\x=-\dfrac{5}{6}-\dfrac{7}{8}\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{-23}{72}\\x=\dfrac{-103}{72}\end{array} \right.\)
Vậy `x={(-23)/72;(-103)/72}`
`b)|x-6/7|=2 3/4`
`=>|x-6/7|= (2.4+3)/4`
`=>|x-6/7|=11/4`
`=>` \(\left[ \begin{array}{l}x-\dfrac{6}{7}=\dfrac{11}{4}\\x-\dfrac{6}{7}=-\dfrac{11}{4}\end{array}\right.\)
`=>` $\left[ \begin{array}{l}x=\dfrac{11}{4}+\dfrac{6}{7}\\x=-\dfrac{11}{4}+\dfrac{6}{7}\end{array}\right.$
`=>` \(\left[ \begin{array}{l}x=\dfrac{101}{28}\\x=\dfrac{-53}{28}\end{array} \right.\)
Vậy `x={101/28;(-53)/28}`
`c)|-1/2+2x|=3/5`
`=>` \(\left[ \begin{array}{l}-\dfrac{1}{2}+2x=\dfrac{3}{5}\\-\dfrac{1}{2}+2x=-\dfrac{3}{5}\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}2x=\dfrac{5}{4}\\2x=-\dfrac{1}{4}\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=\dfrac{5}{8}\\x=\dfrac{-1}{8}\end{array} \right.\)
Vậy `x={5/8;(-1)/8}`
`d)|1 2/3-3x|-1/2=3/4`
`=>|(1.3+2)/3-3x|=3/4+1/2`
`=>|5/3-3x|=5/4`
`=>` $\left[ \begin{array}{l}\dfrac{5}{3}-3x=\dfrac{5}{4}\\\dfrac{5}{3}-3x=-\dfrac{5}{4}\end{array}\right.$
`=>` $\left[ \begin{array}{l}3x=\dfrac{1}{12}\\3x=\dfrac{31}{12}\end{array} \right.$
`=>` \(\left[ \begin{array}{l}x=\dfrac{1}{36}\\x=\dfrac{31}{36}\end{array} \right.\)
Vậy `x={1/36;31/36}`
Bài `3` : Tìm $GTNN$ của các biểu thức :
`a,N=|x-5/8|` `(ĐK:x-5/8∈N)`
Để `N` có $GTNN$ :
`=>x-5/8=0`
`=>x=0+5/8=5/8`
Vậy `N` có $GTNN$ `=0` khi `x=5/8`
`b,N=|x+2 1/7|+12` `(ĐK:x+2 1/7∈N)`
Ta có : `x+15+12=0>=12∀x`
Để `N` có $GTNN$ :
`=>x+2 1/7+12 = 12`
`=>x+(2.7+1)/7=0`
`=>x+15/7=0`
`=>x=(-15)/7`
Vậy `N` có $GTNN$ `=12` khi `x=(-15)/7`
