Đáp án:
`(1-x^2)/x . (x^2/(x+3)-1)+(3x^2-14x+3)/(x^2+3x)<0`
Giải thích các bước giải:
`(1-x^2)/x . (x^2/(x+3)-1)+(3x^2-14x+3)/(x^2+3x)`
`=(1-x^2)/x.(x^2-(x+3))/(x+3)+(3x^2-14x+3)/(x(x+3))`
`=(1-x^2)/x . (x^2-x-3)/(x+3)+(3x^2-14x+3)/(x(x+3))`
`=((1-x^2)(x^2-x-3))/(x(x+3))+(3x^2-14x+3)/(x(x+3))`
`=(-x^4+x^3+4x^2-x-3+3x^2-14x+3)/(x(x+3))`
`=(-x^4+x^3+7x^2-15x)/(x(x+3))`
`=(-x(x^3-x^2-7x+15))/(x(x+3))`
`=-(x^3-4x^2+5x+3x^2-12x+15)/(x+3)`
`=-(x(x^2-4x+5)+3(x^2-4x+5))/(x+3)`
`=-((x+3)(x^2-4x+5))/(x+3)`
`=-(x^2-4x+5)`
`=-(x^2-4x+4+1)`
`=-[(x-2)^2+1]`
`=-(x-2)^2-1`
Vì `(x-2)^2>=0`
`to -(x-2)^2<=0`
`to -(x-2)^2-1<0`
Vậy `(1-x^2)/x . (x^2/(x+3)-1)+(3x^2-14x+3)/(x^2+3x)<0`
