Question

Giai phuong trinh:

\(x^4+\sqrt{x^2+2015}=2015\)

Answer 1

=> \(x^4-2015+\sqrt{x^2+2015}=0\)

<=> \(x^4-\left(x^2+2015\right)+x^2+\sqrt{x^2+2015}=0\)

<=> \(\left(x^2+\sqrt{x^2+2015}\right).\left(x^2-\sqrt{x^2+2015}\right)+\left(x^2+\sqrt{x^2+2015}\right)=0\)

<=> \(\left(x^2+\sqrt{x^2+2015}\right).\left(x^2-\sqrt{x^2+2015}+1\right)=0\)

=> \(x^2-\sqrt{x^2+2015}+1=0\)   (*) (Vì \(x^2+\sqrt{x^2+2015}>0\) với mọi x )

Đặt \(\sqrt{x^2+2015}=t\Rightarrow x^2+2015=t^2\Rightarrow x^2=t^2-2015\)

thay vào (*) ta được: t² - 2015 - t + 1 = 0

=> t² - t - 2014  = 0 

\(\Delta\) = 1 +  4. 2014 = 8057 

=> \(t_1=\frac{1+\sqrt{8057}}{2};t_2=\frac{1-\sqrt{8057}}{2}\)

nhận t₁ => x² = \(\left(\frac{1+\sqrt{8057}}{2}\right)^2-2015\) => x = .....