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TH1: \(x\ge3\)
=>x-3+x+3=4
=>2x=4
=>x=2 (loại)
TH2: \(x<-3\)
=>3-x-x-3=4
=>-2x=4
=>x=-2(loại)
TH3: \(-3\le x<3\)
=>3-x+x+3=4
=>6=4(vô lí)
Vậy không có giá trị nào của x thỏa mãn lx-3l+lx+3l=4
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\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Rightarrow-7< x\le-3\)
Nếu x ∈ Z thì:
\(x\in\left\{-6;-5;-4;-3\right\}\)
\(\dfrac{-28}{4}< x\le\dfrac{-21}{7}\)
\(\Leftrightarrow-4< x\le-3\)
Nếu x nguyên thì x = -3
![](https://rs.olm.vn/images/avt/0.png?1311)
`1 / 3 + 2 / 3 : x = -7`
`2 / 3 : x = -7 - 1 / 3`
`2 / 3 : x = -22 / 3`
`x = 2 / 3 : -22 / 3`
`x = -1 / 11`
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11: |2x-3|-1/3=0
=>|2x-3|=1/3
=>\(\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{10}{3}\\2x=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
12: \(\dfrac{5}{6}-\left|x+\dfrac{1}{4}\right|=\dfrac{1}{4}\)
=>\(\left|x+\dfrac{1}{4}\right|=\dfrac{5}{6}-\dfrac{1}{4}=\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{7}{12}\)
=>\(\left[{}\begin{matrix}x+\dfrac{1}{4}=\dfrac{7}{12}\\x+\dfrac{1}{4}=-\dfrac{7}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{11}{12}\end{matrix}\right.\)
13: \(\left|x-1\right|-2x=\dfrac{1}{2}\)
=>\(\left|x-1\right|=2x+\dfrac{1}{2}\)
=>\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(2x+\dfrac{1}{2}-x+1\right)\left(2x+\dfrac{1}{2}+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(x+\dfrac{3}{2}\right)\left(3x-\dfrac{1}{2}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
14: \(3x-\left|x+15\right|=\dfrac{5}{4}\)
=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16.25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\)
=>\(x=8.125\)
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A = \(\dfrac{5}{9}\cdot\left(\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
=\(\dfrac{5}{9}\cdot\dfrac{5}{7}=\dfrac{25}{63}\)
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b) 128 - 3 . ( x + 4 ) = 23
3 . ( x + 4 ) = 128 - 23
3 . ( x + 4 ) = 105
x + 4 = 105 : 3
x + 4 = 35
x = 35 - 4
x = 31
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Ix+3I=7
\(TH1:x\ge0\Leftrightarrow x\ge-3.\)
\(x+3=7\)
\(x=7-3=4\)
\(TH2:x< 0\Leftrightarrow x< -3\)
\(-\left(x+3\right)=7\)
\(-x-3=7\)
\(-x=10\)
\(x=-10\)
Vậy \(x=\orbr{\begin{cases}4\\-10\end{cases}}\)
[x+2]=7
x=7-3
x=4