`x^3-9x^2+27x-19=0`
`⇔ x^3 - x^2 - 8x^2 + 8x + 19x - 19 =0`
`⇔ x^2(x-1) -8x(x-1) + 19(x-1)=0`
`⇔(x-1)(x^2-8x+19)=0`
`⇔x-1=0` ( do `x^2-8x+19 = x^2 + 8x + 16 + 3= (x+4)^2 + 3 \ge 3 > 0` )
`⇔x=0+1=1.`
Vậy `x=1.`
Ta có HĐT: `a^2+b^2+c^2+2(ab+bc+ca)=(a+b+c)^2.`
`2x^4+3x^3+8x^2+6x+5=0`
`⇔2(2x^4+3x^3+8x^2+6x+5)=0`
`⇔4x^4 + 6x^3 + 16x^2 + 12x + 10=0`
`⇔(x^4+9x^2+4+6x^3+4x^2+12x) + (3x^4 + 3x^2 + 6) =0`
`⇔[(x^2)^2+(3x)^2+2^2+2(x^2. 3x + 3x.2 + 2x^2)] + 3(x^4 + 2. 1/2x^2 + 1/4) + 21/4 =0`
`⇔(x^2+3x+2)^2 + 3(x^2+ 1/2)^2+ 21/4 =0`
Có: `(x^2+3x+2)^2\ge0, (x^2+ 1/2)^2\ge0.`
`⇒(x^2+3x+2)^2 + 3(x^2+ 1/2)^2+ 21/4\ge 0 + 0 + 21/4 = 21/4 > 0 `
`⇒` Không có `x` thỏa mãn.
Vậy phương trình vô nghiệm.
