Question

Giải bất phương trình: \(\sqrt{2\left(x^2-1\right)}\text{ ≤}x+1\left(\text{*}\right)\)

@phynit giúp e

Answer 1

\(\sqrt{2\left(x^2-1\right)}\le x+1\)(ĐKXĐ: \(\left[\begin{matrix}x\ge1\\x=-1\end{matrix}\right.\))

Bình phương cả 2 vế ta được:

2(x^2 - 1) \(\le\)(x + 1)^2

<=> 2x^2 - 2 \(\le\)x^2 + 1 + 2x

<=> 2x^2 - x^2 - 2x \(\le\)1 + 2

<=> x^2 - 2x + 1 \(\le\)4

<=> (x - 1)^2\(\le\) 4

<=> -2 \(\le\)x - 1\(\le\) 2

<=> -1\(\le\) x\(\le\) 3

Vậy \(\left[\begin{matrix}x=-1\\1\le x\le3\end{matrix}\right.\)

\(\le\)x \(\le\)