[Toán 9] Giải phương trình

K

kakashi_hatake

ĐK: $0 \le x \le 1$

$(\sqrt{x+3}-\sqrt{x})(\sqrt{1-x}+1)=1 \\ \leftrightarrow \sqrt{1-x}+1=\dfrac{1}{\sqrt{x+3}-\sqrt{x}}=\dfrac{\sqrt{x+3}+\sqrt{x}}{3} \\ 0 \le x \le 1 \rightarrow \sqrt{1-x}+1 \ge 1 \\ \dfrac{\sqrt{x+3}+\sqrt{x}}{3} \le 1 $

Dấu bằng xảy ra khi x=1

Vậy x=1 thoả mãn
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom