Đáp án:
`a)x(x+2)+x^2=-2x`
`<=>x(x+2)+x^2+2x=0`
`<=>x(x+2)+x(x+2)=0`
`<=>2x(x+2)=0`
`<=>x(x+2)=0`
`<=>[(x=0),(x+2=0):}`
`<=>[(x=0),(x=-2):}`
Vậy `S={0;-2}.`
`b)(x+1)(x^2+4)=x^2+x`
`<=>(x+1)(x^2+4)=x(x+1)`
`<=>(x+1)(x^2-x+4)=0`
Vì `x^2-x+4`
`=(x-1/2)^2+15/4>=15/4>0`
`<=>x+1=0`
`<=>x=-1`.
Vậy `S={-1}.`
`c)x^2(x-5)+5-x=0`
`<=>x^2(x-5)-(x-5)=0`
`<=>(x-5)(x^2-1)=0`
`<=>(x-5)(x-1)(x+1)=0`
`<=>[(x-5=0),(x-1=0),(x+1=0):}`
`<=>[(x=5),(x=1),(x=-1):}`
Vậy `S={5;1;-1}.`
`d)3x^4-9x^3=-9x^2+27x`
`<=>3x^4-9x^3+9x^2-27x=0`
`<=>x^4-3x^3+3x^2-9x=0`
`<=>x(x^3-3x^2+3x-9)=0`
`<=>x[x^2(x-3)+3(x-3)]=0`
`<=>x(x-3)(x^2+3)=0`
Vì `x^2+3>=3>0AAx`
`=>x(x-3)=0`
`<=>[(x=0),(x-3=0):}`
`<=>[(x=0),(x=3):}`
Vậy `S={0;3}.`
