giải pt (x^2 + 2x – 5)^2 = (x^2 – x + 5)^2 x(x + l)(x + 2)(x + 3) = 8;

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1. Giải thích các bước giải:

   a/ $(x^2+2x-5)^2=(x^2-x+5)^2$

   $⇔ (x^2+2x-5)^2-(x^2-x+5)^2=0$

   $⇔ (x^2+2x-5-x^2+x-5)(x^2+2x-5+x^2-x+5)=0$

   $⇔ (3x-10)(2x^2+x)=0$

   $⇔ (3x-10)x(2x+1)=0$

   $⇔ \left[ \begin{array}{l}3x-10=0\\x=0\\2x+1=0\end{array} \right.$

   $⇔ \left[ \begin{array}{l}x=\dfrac{10}{3}\\x=0\\x=-\dfrac{1}{2}\end{array} \right.$

   $\text{Vậy pt có nghiệm: $x=\dfrac{10}{3}$; $x=0$; $x=-\dfrac{1}{2}$}$

   b/ $x(x+1)(x+2)(x+3)=8$

   $⇔ x(x+3)(x+1)(x+2)=8$

   $⇔ (x^2+3x)(x^2+3x+2)=8$

   $\text{Đặt: $t=x^2+3x$}$

   $⇒ t(t+2)=8$

   $⇔ t^2+2t-8=0$

   $⇔ t^2+4t-2t-8=0$

   $⇔ (t+4)(t-2)=0$

   $⇔ \left[ \begin{array}{l}t=-4\\t=2\end{array} \right.$

   $⇔ \left[ \begin{array}{l}x^2+3x+4=0 \text{(pt vô nghiệm)}\\x^2+3x-2=0\end{array} \right.$

   $⇒ x^2+3x-2=0$

   $⇔ 4x^2+12x-8=0$

   $⇔ 4x^2+12x+9-17=0$

   $⇔ (2x+3)^2-17=0$

   $⇔ (2x+3+\sqrt{17})(2x+3-\sqrt{17})=0$

   $⇔ \left[ \begin{array}{l}x=\dfrac{-3-\sqrt{17}}{2}\\x=\dfrac{\sqrt{17}-3}{2}\end{array} \right.$

   $\text{Vậy pt có nghiệm: $x=\dfrac{-3-\sqrt{17}}{2}$; $x=\dfrac{\sqrt{17}-3}{2}$}$

   Chúc bạn học tốt !!!

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