Question

P=\(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}}{x-1}\) kết quả rút gọn \(\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\) a, Tìm x khi P<\(\dfrac{1}{2}\) b, Tính giá trị P Khi x=\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

Answer 1

a,\(P=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x-1}}\)
P<\(\dfrac{1}{2}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{x}-1}< \dfrac{1}{2} \)
\(\Leftrightarrow\dfrac{1}{2}< \dfrac{2}{\sqrt{x}-1}\)\(\Leftrightarrow\dfrac{2}{4}< \dfrac{2}{\sqrt{x}-1}\)
\(\Rightarrow4>\sqrt{x}-1 \Leftrightarrow5>\sqrt{x}\)
\(\Leftrightarrow25>x\)
b, x=\(\sqrt{4+2.2.\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
= \(\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
= \(|2+\sqrt{3}|+|2-\sqrt{3}|\)
= \(2+\sqrt{3}+2-\sqrt{3}=4\)
suy ra P=\(\dfrac{\sqrt{4}-3}{\sqrt{4}-1}=\dfrac{-1}{1}=-1\)