Đáp án+Giải thích các bước giải:
`ĐKXĐ:x\ne 0;x\ne -3`
Ta có:
`A=\frac{1-x^2}{x}.(\frac{x^2}{x+3}-1)+\frac{3x^2-14x+3}{x^2+3x}`
`=\frac{1-x^2}{x}.\frac{x^2-(x+3)}{x+3}+\frac{3x^2-14x+3}{x(x+3)}`
`=\frac{1-x^2}{x}.\frac{x^2-x-3}{x+3}+\frac{3x^2-14x+3}{x(x+3)}`
`=\frac{(1-x^2)(x^2-x-3)}{x(x+3)}.+\frac{3x^2-14x+3}{x(x+3)}`
`=\frac{x^2-x-3-x^4+x^3+3x^2}{x(x+3)}+\frac{3x^2-14x+3}{x(x+3)}`
`=\frac{-x^4+x^3+4x^2-x-3}{x(x+3)}+\frac{3x^2-14x+3}{x(x+3)}`
`=\frac{-x^4+x^3+4x^2-x-3+3x^2-14x+3}{x(x+3)}`
`=\frac{-x^4+x^3+7x^2-15x}{x(x+3)}`
`=\frac{-x(x^3-x^2-7x+15)}{x(x+3)}`
`=\frac{-(x^3-x^2-7x+15)}{x+3}`
`=\frac{-(x^3+3x^2-4x^2-12x+5x+15)}{x+3}`
`=\frac{-[x^2(x+3)-4x(x+3)+5(x+3)]{x+3}`
`=\frac{-(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`
`=>-(x-2)^2<=0`
`=>-(x-2)^2-1<0`
`=>A<0` với mọi `x\ne 0;x\ne -3`
Vậy `A` luôn có giá trị bằng số âm với mọi `x\ne 0;x\ne -3`
