Tìm x biết: a) $|x-1,3|-\frac{1}{4}=2,75$ b) $11:(2x-1)=5:(3x+2)$ c) $(\frac{4}{3}x+1)(\frac{6}{7}x-2)=0$ d) $\frac{x+1}{3}=\frac{27}{x+1}$

Lời giải:

`a, |x - 1,3| - 1/4 = 2,75`

  `|x - 1,3| = 2,75 + 1/4`

  `|x - 1,3| = 3`

$\left[\begin{matrix} x - 1,3 = 3\\ x - 1,3 = -3\end{matrix}\right.$

$\left[\begin{matrix} x = 3 + 1,3\\ x = -3 + 1,3\end{matrix}\right.$

$\left[\begin{matrix} x = 4,3\\ x = -1,7\end{matrix}\right.$

Vậy `x = 4,3` hoặc `x = -1,7`

`b, 11 : (2x - 1) = 5 : (3x + 2)` 

  `11/(2x - 1) = 5/(3x + 2)`

  `11(3x + 2) = 5(2x - 1)`

  `33x + 22 = 10x - 5`

  `33x - 10x = -22 - 5`

  `23x = -27`

  `x = -27 : 23`

  `x = (-27)/23`

Vậy `x = (-27)/23`

`c, (4/3x + 1)(6/7x - 2) = 0`

$\left[\begin{matrix} \dfrac{4}{3}x + 1 = 0\\ \dfrac{6}{7}x - 2 = 0\end{matrix}\right.$

$\left[\begin{matrix} \dfrac{4}{3}x = -1\\ \dfrac{6}{7}x = 2\end{matrix}\right.$

$\left[\begin{matrix} x = \dfrac{-3}{4}\\ x = \dfrac{7}{3}\end{matrix}\right.$

Vậy `x = (-3)/4` hoặc `x = 7/3`

`d, (x + 1)/3 = 27/(x + 1)`

  `(x + 1)(x + 1) = 27 * 3`

  `(x + 1)^2 = 81`

  `(x + 1)^2 = 9^2` hoặc `(x + 1)^2 = (-9)^2`

$\left[\begin{matrix} x + 1 = 9\\ x + 1 = -9\end{matrix}\right.$

$\left[\begin{matrix} x = 8\\ x = -10\end{matrix}\right.$

Vậy `x = 8` hoặc `x = -10`