a) Thay x=4(TMĐK) vào B ta có:
\(B=\dfrac{4-\sqrt{4}}{2\sqrt{4}+1}=\dfrac{2}{5}\)
Vậy x=4 thì B=\(\dfrac{2}{5}\)
b)\(M=A.B\)
M =\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right).\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
M= \(\left(\dfrac{1}{\sqrt{x-1}}+\dfrac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
M= \(\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
M= \(\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
c)\(M=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\)
\(\Leftrightarrow3\sqrt{x}=\sqrt{x}+1\)
\(\Leftrightarrow2\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy với x=\(\dfrac{1}{4}\) thì M=\(\dfrac{1}{3}\)
