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\(2x+\dfrac{1}{2}+4x+\dfrac{1}{6}+6x+\dfrac{1}{6}=12\dfrac{11}{12}\Leftrightarrow\)\(2x+\dfrac{1}{2}+4x+\dfrac{1}{6}+6x+\dfrac{1}{6}=\dfrac{155}{12}\)\(\Leftrightarrow24x+6+48x+2+72x+2=155\)\(\Leftrightarrow24x+48x+72x=155-6-2-2\)\(\Leftrightarrow144x=145\)\(\Leftrightarrow x=\dfrac{145}{144}\)
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\(12:\left(x-1\right)+6:\left(x-1\right)=6\)
\(\Leftrightarrow\left(12+6\right):\left(x-1\right)=6\)
\(\Leftrightarrow18:\left(x-1\right)=6\)
\(\Leftrightarrow x-1=6\cdot18\)
\(\Leftrightarrow x-1=108\)
\(\Leftrightarrow x=109\)
x+10.x+1=122
x.(1+10)+1=122
x.(1+10)=122-1
x.11=121
x=121:11
x=11
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a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran
\(\text{|x-1|+|1-x|=6}\)
\(\text{|x-1|+|x-1|=6}\)
\(\text{2|x-1|=6}\)
\(\text{|x-1|=3}\)
\(x-1=\pm3\)
\(x\in\left\{-2;4\right\}\)