Bạn tham khảo lời giải tại đây:
https://hoc24.vn/cau-hoi/.7125804233195
a)B = \(\left(\dfrac{1}{\sqrt{y}\left(\sqrt{y}+2\right)}-\dfrac{\sqrt{y}}{\sqrt{y}\left(\sqrt{y}+2\right)}\right).\dfrac{\left(\sqrt{y}+2\right)^2}{1-\sqrt{y}}\)
= \(\dfrac{1-\sqrt{y}}{\sqrt{y}\left(\sqrt{y}+2\right)}\). \(\dfrac{\left(\sqrt{y}+2\right)^2}{1-\sqrt{y}}\)= \(\dfrac{\sqrt{y}+2}{\sqrt{y}}\)
b) B=\(\dfrac{5}{3}\Leftrightarrow\dfrac{\sqrt{y}+2}{\sqrt{y}}=\dfrac{5}{3}\)
\(\Leftrightarrow\)\(3\sqrt{y}+6=5\sqrt{y}\)
\(\Leftrightarrow\)\(2\sqrt{y}=6\)\(\Rightarrow\)\(\sqrt{y}=3\)\(\Rightarrow y=9\)
Vậy để B = \(\dfrac{5}{3}\)thì y = 9
