Question

cho bt: p=\(\left(\dfrac{\sqrt{y}}{\sqrt{y}-2}+\dfrac{\sqrt{y}}{\sqrt{y}+2}\right)\).\(\dfrac{y-4}{\sqrt{4y}}\)\

a) rút gọn p

b) tìm giá trị của y để p> 3

Answer 1

\(P=\left(\dfrac{\sqrt{y}}{\sqrt{y}-2}+\dfrac{\sqrt{y}}{\sqrt{y}+2}\right)\cdot\dfrac{y-4}{\sqrt{4y}}\)(1)

ĐKXĐ \(y>0;y\ne4\)

a) pt (1) <=> \(\left(\dfrac{\sqrt{y}\cdot\left(\sqrt{y}+2\right)+\sqrt{y}\cdot\left(\sqrt{y}-2\right)}{y-4}\right)\cdot\dfrac{y-4}{\sqrt{4y}}\)

\(\Leftrightarrow\dfrac{\sqrt{y}\cdot\left(y-4\right)}{\sqrt{4y}}\)

\(\Leftrightarrow\dfrac{\sqrt{y^3}-4\sqrt{y}}{2\sqrt{y}}\)

\(\Leftrightarrow\dfrac{y}{2}-2\)

b) Để \(P>3\Leftrightarrow\dfrac{y}{2}-2>3\)

\(\Leftrightarrow\dfrac{y}{2}>5\Rightarrow y>10\)

Vậy để P > 3 thì y > 10.