Bài 2 (2 điểm). Tìm x, biết: a) $\frac{2}{3}$ - 3x = 2 b) $\frac{x + 1}{5}$ = $\frac{3}{4}$ c) $(x - 2)^{3}$ : (-3) = 9 d) $\frac{3}{4}$ - |x - $\frac{3}{2}$| = $\frac{1}{6}$

Tham khảo :

`a .  ` `2/3-3x=2`

`=>` `3x=2/3-2`

`=>` `3x=-4/3`

`=>` `x=-4/9`

Vậy `x=-4/9`

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`b . ` `(x+1)/5=3/4`

`=>` `4(x+1)=15`

`=>` `(x+1)=15/4`

`=>` `x+1=15/4`

`=>` `x=15/4-1`

`=>` `x=11/4`

Vậy `x=11/4`

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`c . ` `(x-2)^3:(-3)=9`

`=>` `(x-2)^3=-27`

`=>` `(x-2)^3=(-3)^3`

`=>` `(x-2)=(-3)`

`=>` `x-2=(-3)`

`=>` `x=(-3)+2`

`=>` `x=-1`

Vậy `x=-1`

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`d . ` `3/4-|x-3/2|=1/6`

`=>` `|x-3/2|=3/4-1/6`

`=>` `|x-3/2|=7/12`

`=>` $\left[\begin{matrix} x-\dfrac{3}{2}=\dfrac{7}{12}\\ x-\dfrac{3}{2}=-\dfrac{7}{12}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=\dfrac{7}{12}+\dfrac{3}{2} \\ x=\dfrac{7}{12}+(-\dfrac{3}{2})\end{matrix}\right.$

`=>` $\left[\begin{matrix} x=\dfrac{25}{12}\\ x=-\dfrac{11}{12}\end{matrix}\right.$

Vậy `x \in {25/12;-11/12}`