[Toán 9] CHUYÊN ĐỀ Max, Min

I

iceghost

$P=\dfrac{x + 3\sqrt{x-1} + 1}{x + 4\sqrt{x-1} + 2} \\
=\dfrac{x+4\sqrt{x-1}+2-\sqrt{x-1}-1}{x + 4\sqrt{x-1} + 2} \\
=1-\dfrac{\sqrt{x-1}+1}{x + 4\sqrt{x-1} + 2} \\
=1-\dfrac{\sqrt{x-1}+1}{(x-1) + 4\sqrt{x-1} + 4-1} \\
=1-\dfrac{\sqrt{x-1}+1}{(\sqrt{x-1}+2)^2-1} \\
=1-\dfrac{\sqrt{x-1}+1}{(\sqrt{x-1}+1)(\sqrt{x-1}+3)} \\
=1-\dfrac{1}{\sqrt{x-1}+3}$

Ta có : $\sqrt{x-1}+3 \ge 3 \\
\implies \dfrac{1}{\sqrt{x-1}+3} \le \dfrac13 \\
\implies 1-\dfrac{1}{\sqrt{x-1}+3}=P \ge 1-\dfrac13 = \dfrac23$
Dấu '=' xảy ra khi $\sqrt{x-1}=0 \iff x=1$

Có gì sai mong mọi người chỉ giáo :D
 
You must log in or register to reply here.
Top Bottom