Helpppppppppppp! Mình cần gấp Chứng minh $\frac{4\sqrt{x}+4}{x+2\sqrt{x}+5}$ $\leq$ 1
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$\dfrac{4\sqrt{x} + 4}{x + 2\sqrt{x} + 5}$
`=` $\dfrac{x + 2\sqrt{x} + 5 - x + 2\sqrt{x} - 1}{x + 2\sqrt{x} + 5}$
`=` $\dfrac{x + 2\sqrt{x} + 5 - (x - 2\sqrt{x} + 1)}{x + 2\sqrt{x} + 5}$
`=` $\dfrac{x + 2\sqrt{x} + 5 - (\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$
`=` $\dfrac{x + 2\sqrt{x} + 5}{x + 2\sqrt{x} + 5}$ `-` $\dfrac{(\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$
`= 1 -` $\dfrac{(\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$
Vì: `x + 2\sqrt{x} + 5 = x + 2\sqrt{x} + 1 + 4 =` `(\sqrt{x} + 1)²` `+ 4 > 0`
`(\sqrt{x} - 1)² >= 0`
`=>` $\dfrac{(\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$ `>= 0`
`=>` `-`$\dfrac{(\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$ `≤ 0`
`=> 1 -` $\dfrac{(\sqrt{x} - 1)²}{x + 2\sqrt{x} + 5}$ `≤ 1`
Dấu `=` xảy ra khi: `\sqrt{x} = 1` `⇔ x = 1`
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