a) $\dfrac{x+  6}{5} - \dfrac{x-2}{3} < 2$

$<=> \dfrac{x+6}{5} - \dfrac{x-2}{3} - 2 <0 $
$<=> \dfrac{6(x+6) - 10(x - 2) - 2.30}{30}<0$

$=> 6x + 36 - 10x + 20 - 60 < 0$

$<=> -4x -4 <0$

$<=> 4x > 4$

$<=> x > 1$ 

b) $\dfrac{x+5}{4} - \dfrac{x^2 - 3}{6} \geq 1 - \dfrac{2x^2 -1}{12}$

$<=> $\dfrac{x + 5}{4} - \dfrac{x^2 - 3}{6} - 1 + \dfrac{2x^2 - 1}{12} \geq 0$

$<=> $\dfrac{3(x+5) - 2(x^2 - 3) - 12 +2x^2 - 1}{12} \geq 0$

$=> 3x + 15 - 2x^2 + 6 - 12 +2x^2 - 1 \geq 0$

$<=> 3x + 8 \geq 0$

$<=> x \geq \dfrac{-8}{3}$

c) $\dfrac{1 - 5x}{x-1} \geq 1$

$<=> \dfrac{1 - 5x}{x - 1} - 1 \geq 0$

$<=> \dfrac{1 - 5x - x + 1}{x - 1} \geq 0$

$<=> \dfrac{2 - 6x}{x - 1} \geq 0$
=> \(\left[ \begin{array}{l}\begin{cases} 2 - 6x \geq 0\\ x - 1 > 0\\\end{cases} \\ \begin{cases} 2 - 6x \leq 0 \\x - 1 < 0 \\\end{cases} \\\end{array} \right.\)

<=> \(\left[ \begin{array}{l}\begin{cases} x \leq \dfrac{1}{3} \\ x > 1 \\\end{cases} (loại) \\ \begin{cases} x \geq \dfrac{1}{3} \\x <1 \\\end{cases} \\\end{array} \right.\)

=> $\dfrac{1}{3} \leq x < 1$

d) $x^2 - 4x > 0$

$<=> x(x -  4) > 0$

=> => \(\left[ \begin{array}{l}\begin{cases} x > 0 \\ x - 4 > 0\\\end{cases} \\ \begin{cases} x <0 \\x - 4 < 0 \\\end{cases} \\\end{array} \right.\) 

=> \(\left[ \begin{array}{l}x> 4\\x< 0 \end{array} \right.\) 

Đáp án:

Giải thích các bước giải:

 a) `\frac{x+6}{5}-\frac{x-2}{3}<2`

`⇔ \frac{3(x+6)}{15}-\frac{5(x-2)}{15}-\frac{30}{15}<0`

`⇔ \frac{3x+18-5x+10-30}{15}<0`

`⇔ \frac{-2x-2}{15}<0`

`⇒ -2x-2<0`

`⇔ x> -1`

Vậy `S={x|x> -1}`

b) `\frac{x+5}{4}-\frac{x^2-3}{6} \ge 1-\frac{2x^2-1}{12}`

`⇔ \frac{3(x+5)}{12}-\frac{2(x^2-3)}{12} - \frac{12}{12}+\frac{2x^2-1}{12} \ge 0`

`⇔ \frac{3x+15-2x^2+6-12+2x^2-1}{12} \ge 0`

`⇔ \frac{3x+8}{12} \ge 0`

Vậy `S={x|x > - 8/3}`

c) `\frac{1-5x}{x-1} \ge 1`

`⇔ \frac{1-5x}{x-1}-1 \ge 0`

`⇔ \frac{1-5x-x+1}{x-1} \ge 0`

`⇔ \frac{2-6x}{x-1} \ge 0`

TH1: \(\begin{cases} 2-6x \ge 0\\x-1 > 0\end{cases}\)

`⇔` \(\begin{cases} x \le \dfrac{1}{3}\\x > 1\end{cases}\) (vô lí)

TH2: \(\begin{cases} 2-6x \le 0\\x-1 < 0\end{cases}\)

`⇔` \(\begin{cases} x \ge \dfrac{1}{3}\\x < 1\end{cases}\)

`⇒ 1/3 \le x <1`

Vậy `S={x|1/3 \le x < 1}`

d) `x^2-4x > 0`

`⇔ x(x-4)>0`

TH1: \(\begin{cases} x>0\\x-4>0\end{cases}\)

`⇔` \(\begin{cases} x>0\\x > 4\end{cases}\)

`⇒ x>4`

TH2: \(\begin{cases} x< 0\\x-4 < 0\end{cases}\)

`⇔` \(\begin{cases} x <0\\x < 4\end{cases}\)

`⇒x<0`

Vậy `S={x|` \(\left[ \begin{array}{l}x>4\\x<0\end{array} \right.\)`}`