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

Đáp án:

Giải thích các bước giải:

`\bb a)`

`x + 1 (1)/(2) = -5,6`

`=> x+3/2 = -28/5`

`=> x  = (-28/5) - 3/2`

`=> x = -71/10`

Vậy `x = -71/10`

`\bb b)`

`|x-1/4| = 5/4`

`<=>` $\left[\begin{matrix} x-\dfrac{1}{4}=\dfrac{5}{4}\\ x-\dfrac{1}{4}=-\dfrac{5}{4}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{5}{4} + \dfrac{1}{4}\\ x=-\dfrac{5}{4}+\dfrac{1}{4}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{3}{2} \\ x=-1\end{matrix}\right.$

Vậy `x in {3/2;-1}`

`\bb c)`

`(1/5-3/2x)^(2) = 9/4`

`=> (1/5 - 3/2x)^(2) = (+- 3/2)^(2)`

`<=>`$\left[\begin{matrix} \dfrac{1}{5}-\dfrac{3}{2}x = \dfrac{3}{2}\\ \dfrac{1}{5}-\dfrac{3}{2}x = -\dfrac{3}{2}\end{matrix}\right.$

`<=>`$\left[\begin{matrix} \dfrac{3}{2}x = -\dfrac{13}{10}\\ \dfrac{3}{2}x = \dfrac{17}{10}\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x = -\dfrac{13}{15}\\ x = \dfrac{17}{15}\end{matrix}\right.$

Vậy `x in {-13/15;17/15}`