a,
ĐKXĐ: `x>0;x\ne1`
$\\$
b,
`K=(15sqrtx-11)/(x+2sqrtx-3)+(3sqrtx)/(1-sqrtx)-(2sqrtx+3)/(sqrtx+3)`
`=(15sqrtx-11)/((sqrtx-1)(sqrtx+3))-(3sqrtx)/(sqrtx-1)-(2sqrtx+3)/(sqrtx+3)`
`=(15sqrtx-11)/((sqrtx-1)(sqrtx+3))-(3sqrtx(sqrtx+3))/((sqrtx-1)(sqrtx+3))-((2sqrtx+3)(sqrtx-1))/((sqrtx-1)(sqrtx+3))`
`=(15sqrtx-11-3x-9sqrtx-2x+2sqrtx-3sqrtx+3)/((sqrtx-1)(sqrtx+3))`
`=(-5x+5sqrtx-8)/((sqrtx-1)(sqrtx+3))=(-5x+5sqrtx-8)/(x+2sqrtx-3)`
$\\$
c,
Ta có: `K=1/2`
`<=>(-5x+5sqrtx-8)/(x+2sqrtx-3)=1/2`
`<=>2(-5x+5sqrtx-8)=x+2sqrtx-3`
`<=>-10x+10sqrtx-16=x+2sqrtx-3`
`<=>11x-8sqrtx+13=0`
`<=>11x-8sqrtx+16/11-16/11+13=0`
`<=>(sqrt(11x)-4/(sqrt11))^2+127/11>0` với mọi `x`
`=>` Phương trình vô nghiệm
