Question

Giải phương trình:

\(\sqrt{8+x^2}+\sqrt{2-x^2}=4\)

Answer 1

`\sqrt{8+x^2}+\sqrt{2-x^2}=4`         `ĐK: -\sqrt{2} <= x <= \sqrt{2}`

`<=>8+x^2+2-x^2+2\sqrt{(8+x^2)(2-x^2)}=16`

`<=>2\sqrt{16-8x^2+2x^2-x^4}=6`

`<=>\sqrt{-x^4-6x^2+16}=3`

`<=>-x^4-6x^2+16=9`

`<=>x^4+6x^2-7=0`

`<=>x^4+7x^2-x^2-7=0`

`<=>(x^2+7)(x^2-1)=0`

`<=>` $\left[\begin{matrix} x^2=-7\text{ (Vô lí)}\\ x^2=1\end{matrix}\right.$

`<=>x=+-1`  (t/m)

Vậy `S={+-1}`