Đáp án:
Giải thích các bước giải:
Bài 1:
a) `\frac{5x-2}{3}+x=1+\frac{5-3x}{2}`
`⇔ \frac{2.(5x-2)}{6}+\frac{6x}{6}=\frac{6}{6}+\frac{3.(5-3x)}{6}`
`⇔ 2(5x-2)+6x=6+3(5-3x)`
`⇔ 10x-4+6x=6+15-9x`
`⇔ 25x=25`
`⇔ x=1`
Vậy `S={1}`
b) `\frac{1}{x-1}=\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}`
`ĐK: x \ne 1`
`⇔ \frac{1}{x-1}-\frac{2x^2-5}{x^3-1}-\frac{4}{x^2+x+1}=0`
`⇔ \frac{x^2+x+1}{(x-1)(x^2+x+1)}-\frac{2x^2-5}{(x-1)(x^2+x+1)}-\frac{4(x-1)}{(x-1)(x^2+x+1)}`
`⇔ x^2+x+1-2x^2+5-4x+4=0`
`⇔-x^2-3x+10=0`
`⇔ x^2+3x-10=0`
`⇔ (x+5)(x-2)=0`
`⇔` \(\left[ \begin{array}{l}x+5=0\\x-2=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-5\ (TM)\\x=2\ (TM)\end{array} \right.\)
Vậy `S={-5;2}`
Bài 2:
`4-\frac{x+7}{3} \ge \frac{x-6}{2}+5x`
`⇔ \frac{24}{6}-\frac{2(x+7)}{6} \le \frac{3(x-6)}{6}+\frac{30x}{6}`
`⇔ 24-2(x+7) \le 3(x-6)+30x`
`⇔ 24-2x-14 \le 3x-18+30x`
`⇔ -2x-3x-30x \le -24+14-18`
`⇔ -35x \le -28`
`⇔ x \ge \frac{4}{5}`
Vậy `S=[\frac{4}{5};+∞)`
