Punctul e)
[tex] \dfrac{x+1}{2} + |-\tfrac{3}{4} | = \dfrac{x}{8} \\ \dfrac{x+1}{2} +\dfrac{3}{4} = \dfrac{x}{8} \\ 4x+4 + 6 = x \\ 3x=-10 \Rightarrow \tt x=-\dfrac{10}{3} [/tex]
Punctul f)
[tex] 3 \left( x+\dfrac{1}{3} \right) = 2 \left( x-\dfrac{1}{2} \right) \\ 3x+1 =2x-1 \\ 3x-2x=-1-1 \\ \Rightarrow \tt x=-2 [/tex]
Punctul g)
[tex] \dfrac{1}{2} \left[ \left( \dfrac{1}{3} -x \right) +\dfrac{1}{4} \right] = \dfrac{1}{24} \\ \left( \dfrac{1}{3} -x \right) +\dfrac{1}{4} =\dfrac{1}{12} \\ \dfrac{1}{3} -x = -\dfrac{1}{6} \\ x= \dfrac{1}{3} + \dfrac{1}{6} \\ \tt x = \dfrac{1}{2} [/tex]
Punctul h)
[tex] \dfrac{1}{9} \left[ \left( \dfrac{1}{8} x -1\right) -1 \right] =0 \\ \left( \dfrac{1}{8} x -1\right) -1 =0 \\ \dfrac{1}{8} x -1=1 \\ \dfrac{1}{8} x =2 \Rightarrow \tt x=16 [/tex]
Punctul i)
[tex] \dfrac{0,9-x}{0,6}-\dfrac{3-x}{2} = \dfrac{x-2,7}{0,3} \\ \dfrac{1,8-2x-1,8+0,6x}{1,2} =\dfrac{x-2,7}{0,3} \\ \dfrac{-1,4x}{1,2}=\dfrac{x-2,7}{0,3} \\ \dfrac{-0,7x}{0,6} = \dfrac{x-2,7}{0,3} \\ -0,7x= 2x-5,4 \\ 2,7x=5,4 \Rightarrow \tt x=2 [/tex]
Punctul j)
[tex] \left| x-\dfrac{3}{4} \right| = \dfrac{1}{8} \\ x-\dfrac{3}{4}=\dfrac{1}{8} \ \ sau \ \ x-\dfrac{3}{4} =-\dfrac{1}{8} \\ x= \dfrac{1}{8} + \dfrac{3}{4} \ \ sau \ \ x= \dfrac{3}{4} - \dfrac{1}{8} \\ \tt x= \dfrac{7}{8} \ \ sau \ \ x=\dfrac{5}{8} [/tex]
