giải pt 8(x+1/x)^2 +4(x^2+1/x)^2 -4(x^2+1/x^2)(x+1/x)^2 =(x+4)^2 cần gấp lắm ạ

0 bình luận về “giải pt 8(x+1/x)^2 +4(x^2+1/x)^2 -4(x^2+1/x^2)(x+1/x)^2 =(x+4)^2 cần gấp lắm ạ”

1. $8.\bigg(x+\dfrac{1}{x}\bigg)^2+4.\bigg(x^2+\dfrac{1}{x^2}\bigg)^2-4.\bigg(x^2+\dfrac{1}{x^2}\bigg).\bigg(x+\dfrac{1}{x}\bigg)^2 = (x+4)^2$

   $ĐKXĐ : x \neq 0$

   Ta có :

   $8.\bigg(x+\dfrac{1}{x}\bigg)^2+4.\bigg(x^2+\dfrac{1}{x^2}\bigg)^2-4.\bigg(x^2+\dfrac{1}{x^2}\bigg).\bigg(x+\dfrac{1}{x}\bigg)^2 = (x+4)^2$

   $⇔ 8.\bigg(x+\dfrac{1}{x}\bigg)^2+4.\bigg[\bigg(x^2+\dfrac{1}{x^2}+2\bigg)-2\bigg]^2 – 4.\bigg[\bigg(x^2+\dfrac{1}{x^2}+2\bigg)-2\bigg].\bigg(x+\dfrac{1}{x}\bigg)^2= (x+4)^2$

   $⇔ 8.\bigg(x+\dfrac{1}{x}\bigg)^2+4.\bigg[\bigg(x+\dfrac{1}{x}\bigg)^2-2\bigg]^2 – 4.\bigg[\bigg(x+\dfrac{1}{x}\bigg)^2-2\bigg].\bigg(x+\dfrac{1}{x}\bigg)^2= (x+4)^2$

   $⇔ 8.\bigg(x+\dfrac{1}{x}\bigg)^2 + 4.\bigg(x+\dfrac{1}{x}\bigg)^4- 16.\bigg(x+\dfrac{1}{x}\bigg)^2+16 -4.\bigg(x+\dfrac{1}{x}\bigg)^4-8.\bigg(x+\dfrac{1}{x}\bigg)^2 = (x+4)^2$

   $⇔ (x+4)^2=16$

   $⇔ \left[ \begin{array}{l}x+4=4\\x+4=-4\end{array} \right.$ $⇔ \left[ \begin{array}{l}x=0 \text{( Loại)}\\x=-8 \text{( Thỏa mãn )}\end{array} \right.$

   Vậy phương trình đã cho có tập nghiệm $S = \{-8\}$

    

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