a) ĐKXĐ: \(x>0;x\ne4\)

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

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

\(=\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\sqrt{x}\left(\sqrt{x}+1\right)\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}-2}\)

b) Để biểu thức \(Q\) có giá trị âm thì \(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

\(\Rightarrow\sqrt{x}-2< 0\) (vì \(3\sqrt{x}>0\forall x>0;x\ne4\))

\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0\le x< 4\) 

Kết hợp với điều kiện xác định của \(x\), ta được: \(0< x< 4\)

\(\text{#}\mathit{Toru}\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)

\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)

a) ĐKXĐ: \(x\ge0;x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

Giúp mình với

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

\(=\dfrac{-2\sqrt{x}}{2\left(\sqrt{x}-1\right)}=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

ĐKXĐ: \(x>0;x\ne1\)

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

\(=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(x=36\Rightarrow A=\dfrac{\sqrt{36}-1}{\sqrt{36}}=\dfrac{5}{6}\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)

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

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

a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)

a: Thay x=36 vào B, ta được:

\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)