[tex]\sqrt{2x^{2}-8x+6}+ \sqrt{4x-x^{2}-3}\ \textless \ x-1[/tex]

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вопрос

To solve this inequality, we first need to simplify the expressions inside the square roots:

For the first square root expression
[tex]2x^{2} - 8x + 6 = 2(x^{2} - 4x + 3) = 2(x - 3)(x - 1)[/tex]

For the second square root expression
[tex]4x - x^{2} - 3 = -x^{2} + 4x - 3 = -(x^{2} - 4x + 3) = -((x - 3)(x - 1))[/tex]

Now the inequality becomes
[tex]\sqrt{2(x - 3)(x - 1)} + \sqrt{-((x - 3)(x - 1))} < x - 1[/tex]

Notice that both square roots have a common factor of [tex]\sqrt{x - 3}[/tex], so let's substitute [tex]u = \sqrt{x-3}[/tex] and simplify the inequalities:

[tex]2u|u| < (u^2+2)u^2[/tex]

Now, we'll split it into two cases:

Case 1: u >
[tex]2u^2 < (u^2+2)u^2[/tex
[tex]2 < u^2[/tex
[tex]u > \sqrt{2}[/tex
But since we started with u > 0, this condition holds in this case.

Case 2: u <
In this case, u becomes -u, so
[tex]2u|u| < (u^2+2)u^2[/tex] become
[tex]-2u^2 < (u^2+2)u^2[/tex
[tex]u^2 > -2[/tex
Which is always true for real numbers.

Therefore, the solution to the inequality is
[tex]u > \sqrt{2} \rightarrow \sqrt{x-3} > \sqrt{2} \rightarrow x - 3 > 2 \rightarrow x > 5[/tex]