Bài 1:
`a)`
`2x - 5 ≥ 0`
`<=> x ≥ 5/2`
`b)`
`x - 2sqrt{x - 1} ≥ 0`
`<=> (sqrt{x - 1} - 1)² ≥ 0` với `∀ x ∈ RR`
`=> x ∈ RR`
`c)`
`x² - 4x + 3 ≥ 0`
`<=> (x - 3)(x - 1) ≥ 0`
`<=>` \(\left[ \begin{array}{l}x ≥ 3\\x ≤ 1\end{array} \right.\)
`d)`
`-x² + 10x - 25 ≥ 0`
`<=> -(x - 5)² ≥ 0` (vô lí)
`=> x ∈ ∅`
Bài 2:
`b)`
`B² = (sqrt{4 + \sqrt{10 + 2\sqrt{5}}} + sqrt{4 - \sqrt{10 + 2sqrt{5}}})²`
`= 4 + sqrt{10 + 2\sqrt{5}} + 2.sqrt{(4 + \sqrt{10 + 2\sqrt{5}}).(4 - \sqrt{10 + 2\sqrt{5}})} + 4 - sqrt{10 + 2\sqrt{5}}`
`= 8 + 2.sqrt{6 - 2\sqrt{5}}`
`= 8 + 2.sqrt{(\sqrt{5} - 1)²}`
`= 8 + 2.(sqrt{5} - 1)`
`= 8 + 2sqrt{5} - 2`
`= 6 + 2sqrt{5}`
`=> B = sqrt{6 + 2\sqrt{5}} = sqrt{(\sqrt{5} + 1)²} = sqrt{5} + 1`
Vậy `B = sqrt{5} + 1`
