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

a) ĐKXĐ: \(x>0;x\ne4\)

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

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

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

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

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

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: \(0< x< 4\)

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

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

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

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

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

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

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

\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)

a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)

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

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

b) \(\dfrac{P}{A}\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)

Vậy \(S=\varnothing\)

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

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

b: Để A>-6 thì \(2\sqrt{x}< 6\)

=>0<x<9 

Kết hợp ĐKXĐ, ta được:

0<x<9 và x<>1

\(a,\)

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

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

\(=\dfrac{x-1}{2\sqrt{x}}.\dfrac{-4x}{x-1}\)

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

Vậy \(A=-2\sqrt{x}\)

\(b,\)Đề \(A\ge-6\) thì \(-2\sqrt{x}\ge-6\) \(\Leftrightarrow\sqrt{x}\le3\) \(\Leftrightarrow x\le3^2\Leftrightarrow x\le9\)

Vậy \(x\le9\) thì \(A\ge-6\)

Bài 1:

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

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

2) \(VT=\dfrac{\left(6a+1\right)\left(a+6\right)+\left(6a-1\right)\left(a-6\right)}{a\left(a-6\right)\left(a+6\right)}.\dfrac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\dfrac{12a^2+12}{a\left(a^2+1\right)}=\dfrac{12\left(a^2+1\right)}{a\left(a^2+1\right)}=\dfrac{12}{a}=VP\)