Toán Lớp 9: Giải pt:
2×4 – 5×3 + 6×2 – 5x + 2 = 0 ( những số cạnh x là số mũ nhé)
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Toán Lớp 9: Giải pt:
2×4 – 5×3 + 6×2 – 5x + 2 = 0 ( những số cạnh x là số mũ nhé)
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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}