Question

cho bt :A = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-1}\)

B=\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\) với \(x\ge0;x\ne1;x\ne4\)

a , tính giá trị của x để P=\(\dfrac{1}{3}\) biết P=B:A

b . So sánh P với \(\dfrac{-2}{3}\)

Answer 1

\(P=B:A\)

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}+3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)

\(P=\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{1}{3}\Leftrightarrow3\sqrt{x}-6=\sqrt{x}+3\)

\(\Leftrightarrow2\sqrt{x}=9\Leftrightarrow\sqrt{x}=4,5\Leftrightarrow x=\dfrac{81}{4}\)

b. \(P=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}=1-\dfrac{5}{\sqrt{x}+3}\)

Ta có: \(-\dfrac{5}{\sqrt{x}+3}\ge-\dfrac{5}{\sqrt{0}+3}=-\dfrac{5}{3}\)

\(\Rightarrow1-\dfrac{5}{\sqrt{x}+3}\ge1-\dfrac{5}{3}=-\dfrac{2}{3}\)

Suy ra: \(P\ge-\dfrac{2}{3}\) khi \(x=0\)