Giải pt: 2x4 - 5x3 + 6x2 - 5x + 2 = 0 ( những số cạnh x là số mũ nhé)
2 câu trả lời
`2x^4-5x^3+6x^2-5x+2=0`
`⇔2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0`
`⇔2x^3(x-1)-3x^2(x-1)+3x(x-1)-2(x-1)=0`
`⇔(x-1)(2x^3-3x^2+3x-2)=0`
`⇔(x-1)(2x^3-2x^2-x^2+x+2x-2)=0`
`⇔(x-1)[2x^2(x-1)-x(x-1)+2(x-1)]=0`
`⇔(x-1)(x-1)(2x^2-x+2)=0`
`⇔(x-1)^2(2x^2-x+2)=0`
⇔\(\left[ \begin{array}{l}(x-1)^2=0 (1)\\2x^2-x+2=0 (2)\end{array} \right.\)
`(1)⇔x-1=0`
`⇔x=1`
`(2)⇔2(x^2-\frac{1}{2}x+1)=0`
`⇔x^2-\frac{1}{2}x+1=0`
`⇔x^2-2x.\frac{1}{4}+(\frac{1}{4})^2+\frac{15}{16}=0`
`⇔(x-\frac{1}{4})^2=\frac{-15}{16}` (Vô lí vì `(x-\frac{1}{4})^2>=0` mà `\frac{-15}{16}<0`)
Vậy ptr đã cho có nghiệm `x=1`
Ta có:
`2x^4 - 5x^3 + 6x^2 - 5x + 2 = 0`
`<=>` `2x^4 - 3x^3 + 3x^2 + 3x^2 - 3x - 2x +2 = 0`
`<=>` `2x^3(x - 1) - 3x^2(x - 1) + 3x(x - 1) - 2(x-1) = 0`
`<=>` `(2x^3 - 3x^2 + 3x - 2)(x - 1) = 0`
`<=>` `(2x^2 - x + 2)(x - 1)(x - 1) = 0`
`<=>` `2(x^2 - 1/2 . x + 1)(x - 1)^2 = 0`
`<=>` `2[(x - 1/2 . x + 1/16) + 15/16](x - 1)^2 = 0`
`<=>` `2[(x - 1/4)^2 + 15/16](x - 1)^2 = 0`
`<=>` $\left[\begin{matrix} 2(x - \dfrac{1}{4} ) + \dfrac{15}{16} = 0 \\ (x - 1)^2 = 0\end{matrix}\right.$
Xét `(x - 1)^2 = 0`
`<=>` `x - 1 = 0`
`<=>` `x = 1`
Xét `2(x - 1/4)^2 + 15/16 = 0`
Vô lí vì `2(x - 1/4)^2 + 15/16 ≥ 15/16` `AA x`
Vậy `S = {1}`