Đáp án: `b)` `3; \frac{-1}{2}`
`c)` `\frac{5}{3}`
`d)` `-1`
Bài 2: `x | x \le (-6)`
Giải thích các bước giải: `b)` `(3x - 1)(x - 3) - 9 + x^2 = 0`
`⇔` `(3x - 1)(x - 3) + (x^2 - 9) = 0`
`⇔` `(3x - 1)(x - 3) + (x - 3)(x + 3) = 0`
`⇔` `(x - 3) (3x - 1 + x + 3) = 0`
`⇔` `(x - 3) (4x + 2) = 0`
`⇔` \(\left[ \begin{array}{l}x-3 = 0\\4x + 2 = 0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = 3\\x = \dfrac{-1}{2}\end{array} \right.\)
`S = {3; \frac{-1}{2}}`
`c)` $\left[\begin{array}{ccc}x - 2\end{array}\right]$ `= 2x - 3`
ĐKXĐ: `2x - 3 \ge = 0`
`⇔` \(\left[ \begin{array}{l}x + 2 = 2x - 3\\-x + 2 = 2x - 3\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x + 2x = -3 + 2\\-x - 2x = -3 - 2\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}-x = 1\\-3x = -5\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1\\x=\dfrac{5}{3}\end{array} \right.\)
`S = {\frac{5}{3}}`
`d)` `\frac{x + 2}{x - 2} - \frac{1}{x} = \frac{2}{x(x - 2)}`
ĐKXĐ: `x\ne 0`; `x\ne 2`
`⇔` `\frac{x(x + 2)}{x(x - 2)} - \frac{1(x - 2)}{x(x - 2)} = \frac{2}{x(x - 2)}`
`⇔` `x(x + 2) - 1(x - 2) = 2`
`⇔` `x^2 + 2x - x + 2 - 2 = 0`
`⇔` `x^2 + x = 0`
`⇔` `x(x + 1) = 0`
`⇔` \(\left[ \begin{array}{l}x=0\\x + 1 = 0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x = 0\\x = -1 \end{array} \right.\)
`S = {-1}`
Bài 2: `\frac{x - 1}{3} - \frac{3x + 5}{2} \ge 1 - \frac{4x + 5}{6}`
`⇔` `\frac{2(x - 1)}{6} - \frac{3(3x + 5)}{6} \ge \frac{6}{6} - \frac{4x + 5}{6}`
`⇔` `2(x - 1) - 3(3x + 5) \ge 6 - (4x + 5)`
`⇔` `2x - 2 - 9x - 15 \ge 6 - 4x - 5`
`⇔` `-7x - 17 \ge 1 - 4x`
`⇔` `-7x + 4x \ge 17 + 1`
`⇔` `-3x \ge 18`
`⇔` `x \le 18 : (-3)`
`⇔` `x\le (-6)`
`S = {x | x \le (-6)}`
