Question

[Kimian Hajan Ruventaren]

Tìm tập nghiệm của bất pt

a) \(2x-\dfrac{x-3}{5}\le4x-1\)

b) \(\sqrt{x^2+2}\le x-1\)

c) \(\sqrt{x-1}+\sqrt{5-x}+\dfrac{1}{x-3}>\dfrac{1}{x-3}\)

Answer 1

[Rimuru tempest]

a) \(2x-\dfrac{x-3}{5}-4x+1\le0\)

\(\Leftrightarrow10x-x+3-20x+5\le0\)

\(\Leftrightarrow-11x+8\le0\)

\(\Leftrightarrow x\ge\dfrac{8}{11}\)

\(\Rightarrow x\in\left(\dfrac{8}{11};+\infty\right)\)

b) \(\sqrt{x^2+2}\le x-1\)

\(\Leftrightarrow x^2+2\le x^2-2x+1\) \(\left(x-1\ge\sqrt{x^2+2}\ge\sqrt{2}\Rightarrow x\ge1+\sqrt{2}\right)\)

\(\Leftrightarrow x\le-\dfrac{1}{2}\)

\(\Rightarrow x\in\varnothing\)

c) \(\sqrt{x-1}+\sqrt{5-x}+\dfrac{1}{x-3}>\dfrac{1}{x-3}\) (\(x\in\left[1;5\right]\backslash\left\{3\right\}\))

\(\Leftrightarrow\sqrt{x-1}+\sqrt{5-x}>0\)

\(\Leftrightarrow4+2\sqrt{\left(x-1\right)\left(5-x\right)}>0\) ( luôn đúng )

vậy \(x\in\left[1;5\right]\backslash\left\{3\right\}\)