Giải phương trình (x^2+5)^2 +(2x^2 +3)^2-(x^2+5)(2x^2+3)-2x^4- 13x-15= 0 17/11/2021 Bởi Alice Giải phương trình (x^2+5)^2 +(2x^2 +3)^2-(x^2+5)(2x^2+3)-2x^4- 13x-15= 0
$(x^2+5)^2 + (2x^2 + 3)^2-(x^2 + 5)(2x^2 + 3)$ – 2x^4 – 13x – 15 = 0 ⇔ $x^4 + 10x^2 + 25 + 4x^4 + 12x^2 + 9 − 2x^4 − 13x^2 − 15 − 2x^4$ − 13x − 15 = 0 ⇔ $x^4 + 9x^2 − 13x + 4 = 0$ ⇔\(\left[ \begin{array}{l}x=0 , 4529981524\\x=0 , 855281464 \end{array} \right.\) Bình luận
$(x^2+5)^2 + (2x^2 + 3)^2-(x^2 + 5)(2x^2 + 3)$ – 2x^4 – 13x – 15 = 0
⇔ $x^4 + 10x^2 + 25 + 4x^4 + 12x^2 + 9 − 2x^4 − 13x^2 − 15 − 2x^4$ − 13x − 15 = 0
⇔ $x^4 + 9x^2 − 13x + 4 = 0$
⇔\(\left[ \begin{array}{l}x=0 , 4529981524\\x=0 , 855281464 \end{array} \right.\)