Tìm x biết
3.|x-2| - |-9| = 12
6(x- 2) - (x-3) = 31
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\(a.\dfrac{3}{2}+\dfrac{-1}{3}< \dfrac{x}{6}< \dfrac{1}{9}+\dfrac{31}{18}\)
\(\Leftrightarrow\dfrac{7}{6}< \dfrac{x}{6}< \dfrac{11}{6}\)
\(\Leftrightarrow7< x< 11\)
\(\Leftrightarrow x\in\left\{8;9;10\right\}\)
\(b.\dfrac{-5}{12}+\dfrac{7}{12}+\dfrac{-1}{12}< \dfrac{x}{12}< \dfrac{2}{15}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{1}{12}< \dfrac{x}{12}< \dfrac{4}{12}\)
\(\Leftrightarrow1< x< 4\)
\(\Leftrightarrow x\in\left\{2;3\right\}\)
a, x + 1/9 - 3/5 = 3/6
x + 1/9 = 3/6 - 3/5
x + 1/9 = -1/10
x = -1/10 - 1/9
x = -19/90
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
\(3\left|x-2\right|-\left|-9\right|=12\\ 3\left|x-2\right|-9=12\\ 3\left|x-2\right|=12+9=21\\ \left|x-2\right|=\dfrac{21}{3}=7\)
Suy ra \(x-2=7\) hoặc \(x-2=-7\)
Vậy \(x=9\) hoặc \(x=-5\)
\(6\left(x-2\right)-\left(x-3\right)=31\\ 6x-12-x+3=31\\ 5x-9=31\\ 5x=31+9=40\\ x=\dfrac{40}{5}=8\)
Vậy \(x=8\)