b. `1-\frac{2x-5}{6}=\frac{3-x}{4}`
`↔\frac{24}{24}-\frac{4(2x-5)}{24}=\frac{6(3-x)}{24}`
`↔24-8x+20=18-6x`
`↔-8x+6x=-24-20+18`
`↔-2x=-26`
`↔x=13`
Vậy: `S=13`
c. `\frac{4}{x-2}-\frac{2}{x-5}+\frac{3x}{(x-2)(x-5)}=0` (ĐKXĐ: `x \ne 2;x \ne 5`)
`↔\frac{4(x-5)}{(x-2)(x-5)}-\frac{2(x-2)}{(x-2)(x-5)}+\frac{3x}{(x-2)(x-5)}=0`
`→4x-20-2x+4+3x=0`
`↔4x-2x+3x=20-4`
`↔5x=16`
`↔x=\frac{16}{5}` (thỏa mãn điều kiện)
Vậy: `S={\frac{16}{5}}`
d. `4x^2+5x-9=0`
`↔(4x^2-4x)+(9x-9)=0`
`↔4x(x-1)+9(x-1)=0`
`↔(x-1)(4x+9)=0`
`↔`\(\left[ \begin{array}{l}x-1=0\\4x+9=0\end{array} \right.\)
`↔`\(\left[ \begin{array}{l}x=1\\x=-\dfrac{9}{4}\end{array} \right.\)
Vậy: `S={1;-\frac{9}{4}}`
