Đáp án:
Giải thích các bước giải:
a) `\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3`
`⇔ \frac{21(4x+3)}{105}-\frac{15(6x-2)}{105}=\frac{35(5x+4)}{105}+\frac{315}{105}`
`⇔ 21(4x+3)-15(6x-2)=35(5x+4)+315`
`⇔ 84x+63-90x+30=175x+140+315`
`⇔ -181x=362`
`⇔ x=-2`
Vậy `S={-2}`
b) `\frac{3}{2x-16}+\frac{3x-20}{x-8}+\frac{1}{8}=\frac{13x-102}{3x-24}`
`ĐK: x \ne 8`
`⇔ \frac{36}{24(x-8)}+\frac{24(3x-20)}{24(x-8)}+\frac{3(x-8)}{24(x-8)}=\frac{8(13x-102)}{24(x-8)}`
`⇔ 36+24(3x-20)+3(x-8)=8(13x-102)`
`⇔ 36+72x-480+3x-24=104x-816`
`⇔ -29x=-348`
`⇔ x=12\ (TM)`
Vậy `S={12}`
c) `\frac{5x-3}{5}+\frac{2x+1}{4} \le \frac{2-3x}{2}-5`
`⇔ \frac{4(5x-3)}{20}+\frac{5(2x+1)}{20} \le \frac{10(2-3x)}{20}-\frac{100}{20}`
`⇔ 4(5x-3)+5(2x+1) \le 10(2-3x)-100`
`⇔ 20x-12+10x+5 \le 20-30x-100`
`⇔ 60x \le -73`
`⇔ x \le -\frac{73}{60}`
Vậy `S=(-∞;-\frac{73}{60}]`
d) `\frac{x}{x-2}+\frac{x+2}{x}>2` ĐK: `x \ne 0,x \ne 2`
`⇔ \frac{x^2}{x(x-2)}+\frac{x^2-4}{x(x-2)}-\frac{2x(x-2)}{x(x-2)}>0`
`⇔ \frac{4x}{x(x-2)}>0`
TH1:\(\begin{cases} 4x>0\\ x^2-2x>0\end{cases}\)
`⇔` \(\begin{cases} x>0\\ \left[ \begin{array}{l}x<0\\x>2\end{array} \right.\end{cases}\)
`⇒ x>2`
TH2: \(\begin{cases} 4x<0\\ x^2-2x<0\end{cases}\)
`⇔` \(\begin{cases} x<0\\0<x<2\end{cases}\) \ (loại)
Vậy `S=(2;+∞)`
