Question

(x²+7)(x²-49)<0

Answer 1

`(x^2+7)(x^2-49)<0`

Vì `x^2+7>=7>0`

`=>x^2-49<0`

`<=>x^2-7x+7x-49<0`

`<=>x(x-7)+7(x-7)<0`

`<=>(x-7)(x+7)<0`

Vì `x+7>x-7`

`=>` $\begin{cases}x+7>0\\x-7<0\\\end{cases}$

`=>` $\begin{cases}x>-7\\x<7\\\end{cases}$

`=>-7<x<7`

Vậy `-7<x<7`

Answer 2

Ta có: \(\left(x^2+7\right)\left(x^2-49\right)< 0\)

mà \(x^2+7>0\)

nên \(x^2-49< 0\)

\(\Leftrightarrow x^2< 49\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>-7\\x< 7\end{matrix}\right.\)

Vậy: -7<x<7