ĐK: x>0,x\(\ne1\)
a) Ta có x=\(\dfrac{1}{4}\) thay vào B thì B=\(\dfrac{4.\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}-1}=\dfrac{1-4}{\dfrac{1}{2}-1}=\dfrac{-3}{-\dfrac{1}{2}}=6\)
Vậy khi x=\(\dfrac{1}{4}\) thì B=6
b) Ta có P=A+B=\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{4x+3\sqrt{x}}{\sqrt{x}}+\dfrac{4x-4}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(4\sqrt{x}+3\right)}{\sqrt{x}}+\dfrac{4\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(4\sqrt{x}+3\right)+\dfrac{4\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\sqrt{x}\left(\sqrt{x}-1\right)-4\sqrt{x}-3+4\left(\sqrt{x}+1\right)=x-\sqrt{x}-4\sqrt{x}-3+4\sqrt{x}+4=x-\sqrt{x}+1\)
Vậy P=\(x-\sqrt{x}+1\)
c) Ta có P=3\(\Leftrightarrow x-\sqrt{x}+1=3\Leftrightarrow x-\sqrt{x}-2=0\Leftrightarrow x-2\sqrt{x}+\sqrt{x}-2=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}+1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=-1\left(loai\right)\end{matrix}\right.\)\(\Leftrightarrow x=4\left(tm\right)\)
Vậy để P=3 thì x=4
