Đáp án:
`10)S={0}`
`11)S={12}`
Giải thích các bước giải:
`10)(x+6)/(x-5)+(x-5)/(x+6)=(2x²+23x+61)/(x²+x-30)(ĐKXĐ:``x`$\neq$ `-6,x`$\neq$ `5)`
`⇔(x+6)/(x-5)+(x-5)/(x+6)=(2x²+23x+61)/[(x-5)(x+6)]`
`⇔[(x+6)²]/[(x-5)(x+6)]+[(x-5)²]/[(x-5)(x+6)]=(2x²+23x+61)/[(x-5)(x+6)]`
`⇒(x+6)²+(x+5)²=2x²+23x+61`
`⇔x²+12x+36+x²+10x+25=2x²+23x+61`
`⇔2x²+22x+61=2x²+23x+61`
`⇔2x²+22x-2x²-23x=61-61`
`⇔-x=0`
`⇔x=0(TM` `ĐKXĐ)`
Vậy `S={0}`
`11)(x+9)/(x²-3x-10)-(x+5)/(x²-25)=1/(x+2)(ĐKXĐ:``x`$\neq$ `5,x`$\neq$ `-5,x`$\neq$ `-2)`
`⇔(x+9)/[(x-5)(x+2)]-(x+5)/(x²-25)=1/(x+2)`
`⇔[(x+9)(x+5)]/[(x²-25)(x+2)]-[(x+5)(x+2)]/[(x²-25)(x+2)]=(x²-25)/[(x²-25)(x+2)]`
`⇔(x+9)(x+5)-(x+5)(x+2)=x²-25`
`⇔x²+5x+9x+45-(x²+2x+5x+10)=x²-25`
`⇔x²+5x+9x+45-x²-2x-5x-10=x²-25`
`⇔7x+35=x²-25`
`⇔-x²+7x+35+25=0`
`⇔-x²+7x+60=0`
`⇔-x²+12x-5x+60=0`
`⇔-x(x-12)-5(x-12)=0`
`⇔(x-12)(-x-5)=0`
`(1)x-12=0⇔x=12(TM` `ĐKXĐ)`
`(2)-x-5=0⇔x=-5(Ko` `TM` `ĐKXĐ)`
Vậy `S={12}`
