Question

CM: \(\dfrac{4\sqrt{x}+4}{x+2\sqrt{x}+5}\le1\)

Answer 1

ĐKXĐ: x ≥ 0

(4√x + 4)/(x + 2√x + 5) ≥ 1

⇔ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

Do x ≥ 0 ⇒ x + 2√x + 5 > 0

⇒ (4√x + 4)/(x + 2√x + 5) - 1 ≤ 0

⇔ (4√x + 4) - (x + 2√x + 5) ≤ 0

⇔ 4√x + 4 - x - 2√x - 5 ≤ 0

⇔ -x + 2√x - 1 ≤ 0

⇔ -(x - 2√x + 1) ≤ 0

⇔ -(√x - 1)² ≤ 0 (luôn đúng)

Vậy (4√x + 4)/(x + 2√x + 5) ≤ 1 với mọi x ∈ R

Answer 2

=>\(\dfrac{4\sqrt{x}+4-1}{x+2\sqrt{x}+5}< =0\)

=>\(\dfrac{4\sqrt{x}+3}{x+2\sqrt{x}+5}< =0\)(vô lý)