`a)(5x+3)/(x²-3x)+(9-x)/(9-3x)(x`$\neq$ `0,x`$\neq$ `3)`
`=(5x+3)/[x(x-3)]+(x-9)/(3x-9)`
`=(5x+3)/[x(x-3)]+(x-9)/[3(x-3)]`
`=[3(5x+3)]/[3x(x-3)]+[x(x-9)]/[3x(x-3)]`
`=[3(5x+3)+x(x-9)]/[3x(x-3)]`
`=(15x+9+x²-9x)/[3x(x-3)]`
`=(x²+6x+9)/[3x(x-3)]`
`b)(6x²+11x+4)/(x³-1)+(2x-1)/(x²+x+1)+7/(1-x)(x`$\neq$ `1)`
`=(6x²+11x+4)/[(x-1)(x²+x+1)]+(2x-1)/(x²+x+1)-7/(x-1)`
`=(6x²+11x+4)/[(x-1)(x²+x+1)]+[(2x-1)(x-1)]/[(x-1)(x²+x+1)]-[7(x²+x+1)]/[(x-1)(x²+x+1)]`
`=[6x²+11x+4+(2x-1)(x-1)-7(x²+x+1)]/[(x-1)(x²+x+1)]`
`=(6x²+11x+4+2x²-2x-x+1-7x²-7x-7)/[(x-1)(x²+x+1)]`
`=[(6x²+2x²-7x²)+(11x-2x-x-7x)+(4+1-7)]/[(x-1)(x²+x+1)]`
`=(x²+x-2)/[(x-1)(x²+x+1)]`
`=(x²+2x-x-2)/[(x-1)(x²+x+1)]`
`=[x(x+2)-(x+2)]/[(x-1)(x²+x+1)]`
`=[(x+2)(x-1)]/[(x-1)(x²+x+1)]`
`=(x+2)/(x²+x+1)`