biểu thức B đâu rồi bạn

a: ĐKXĐ: x>=0; x<>1

b \(A=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

\(=\dfrac{x+2\sqrt{x}-x-\sqrt{x}-1}{x\sqrt{x}-1}\cdot\dfrac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

\(=\dfrac{1}{\sqrt{x}+2}\)

c: Khi x=9-4 căn 5 thì \(A=\dfrac{1}{\sqrt{5}-2+2}=\dfrac{\sqrt{5}}{5}\)

d: căn x+2>=2

=>A<=1/2

Dấu = xảy ra khi x=0

\(\sqrt{x}-\sqrt{2-x}\)

ĐKXĐ: 

\(\left\{{}\begin{matrix}x\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\le2\end{matrix}\right.\)

\(\Leftrightarrow0\le x\le2\)

Vậy \(0\le x\le2\) thì biểu thức có nghĩa 

Đề thiếu rồi. Bạn xem lại.

a: \(A=\sqrt{x}+\dfrac{\sqrt{x}\left(1+2\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}+\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

Khi x=4 thì \(A=2+\dfrac{2\cdot2+1}{2+1}=2+\dfrac{5}{3}=\dfrac{11}{3}\)

b: Khi x=(2-căn 3)^2 thì \(A=2-\sqrt{3}+\dfrac{2\left(2-\sqrt{3}\right)+1}{2-\sqrt{3}+1}\)

\(=2-\sqrt{3}+\dfrac{4-2\sqrt{3}+1}{3-\sqrt{3}}\)

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

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3-\sqrt{3}\right)+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{6-2\sqrt{3}-3\sqrt{3}+3+5-2\sqrt{3}}{3-\sqrt{3}}\)

\(=\dfrac{14-7\sqrt{3}}{3-\sqrt{3}}\)

d: A=2

=>\(\dfrac{x+\sqrt{x}+2\sqrt{x}+1}{\sqrt{x}+1}=2\)

=>\(x+3\sqrt{x}+1=2\left(\sqrt{x}+1\right)=2\sqrt{x}+2\)

=>\(x+\sqrt{x}-1=0\)

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{-1+\sqrt{5}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{-1-\sqrt{5}}{2}\left(loại\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{6-2\sqrt{5}}{4}=\dfrac{3-\sqrt{5}}{2}\)

A đâu em?

\(A=\dfrac{x}{\sqrt{x}+1}+\dfrac{\sqrt{x}+2x}{x+\sqrt{x}}\)