Đáp án:
`e)S={2}`
`f)S={12}`
`g)S=∅`
`h)S={1}`
Giải thích các bước giải:
`e)(x-3)(x²+3x+9)-x(x+4)(x-4)=5`
`⇔(x-3)(x²+x.3+3²)-x(x²-4²)=5`
`⇔x³-3³-x(x²-16)=5`
`⇔x³-27-x³+16x=5`
`⇔(x³-x³)+16x-27=5`
`⇔16x-27=5`
`⇔16x=5+27`
`⇔16x=32`
`⇔x=32:16`
`⇔x=2`
Vậy `S={2}`
`f)(x-2)³-(x+5)(x²-5x+25)+6x²=11`
`⇔x³-3.x².2+3.x.2²-2³-(x+5)(x²-x.5+5²)+6x²=11`
`⇔x³-6x²+12x-8-(x³+5³)+6x²=11`
`⇔x³-6x²+12x-8-(x³+125)+6x²=11`
`⇔x³-6x²+12x-8-x³-125+6x²=11`
`⇔(x³-x³)+(-6x²+6x²)+12x-(8+125)=11`
`⇔12x-133=11`
`⇔12x=11+133`
`⇔12x=144`
`⇔x=144:12`
`⇔x=12`
Vậy `S={12}`
`g)(x+2)(x²-2x+4)(x-2)(x²+2x+4)=-65`
`⇔[(x+2)(x²-2x+4)][(x-2)(x²+2x+4)]=-65`
`⇔[(x+2)(x²-x.2+2²)][(x-2)(x²+x.2+2²)]=-65`
`⇔(x³+2³)(x³-2³)=-65`
`⇔(x^3)^2-(2^3)^2=-65`
`⇔x^6-2^6+65=0`
`⇔x^6-64+65=0`
`⇔x^6+1=0`
Ta có:`x^6≥0∀x`
`⇒x^6+1≥1>0∀x`
`⇒` vô nghiệm
Vậy `S=∅`
`h)(x-1)³+(2-x)(4+2x+x²)+3x(x+2)=16`
`⇔x³-3.x².1+3.x.1²-1³+(2-x)(2²+2.x+x²)+3x²+6x=16`
`⇔x³-3x²+3x-1+2³-x³+3x²+6x=16`
`⇔x³-3x²+3x-1+8-x³+3x²+6x=16`
`⇔(x³-x³)+(-3x²+3x²)+(3x+6x)+(-1+8)=16`
`⇔9x+7=16`
`⇔9x=16-7`
`⇔9x=9`
`⇔x=9:9`
`⇔x=1`
Vậy `S={1}`
