2[tex]log_{2x-8}[/tex]([tex]\sqrt{x+3}-\sqrt{7-x}) \ \textless \ 1[/tex]

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вопрос

To solve this inequality, we first need to simplify the expression inside the logarithm:

[tex]\sqrt{x+3}-\sqrt{7-x}[/tex]

To simplify this expression, we can rewrite it as a single square root term:

[tex]\sqrt{x+3}-\sqrt{7-x} = \frac{(\sqrt{x+3}-\sqrt{7-x})(\sqrt{x+3}+\sqrt{7-x})}{\sqrt{x+3}+\sqrt{7-x}}[/tex]
[tex]= \frac{(x+3) - (7-x)}{\sqrt{x+3}+\sqrt{7-x}}[/tex]
[tex]= \frac{2x-4}{\sqrt{x+3}+\sqrt{7-x}}[/tex]

Now, we can rewrite the given inequality using this expression:

[tex]\log_{2x-8} \left(\frac{2x-4}{\sqrt{x+3}+\sqrt{7-x}}\right) < 1[/tex]

Exponentiating both sides with base tex[/tex]:

[tex]\frac{2x-4}{\sqrt{x+3}+\sqrt{7-x}} < (2x-8)^1[/tex]
[tex]2x-4 < (2x-8)(\sqrt{x+3}+\sqrt{7-x})[/tex]
[tex]2x-4 < 2x(\sqrt{x+3}+\sqrt{7-x}) - 8(\sqrt{x+3}+\sqrt{7-x})[/tex]
[tex]4 < 2x(\sqrt{x+3}+\sqrt{7-x}) - 8(\sqrt{x+3}+\sqrt{7-x})[/tex]

This inequality can be solved further by expanding the terms, simplifying, and finding the values of x that satisfy the inequality.