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

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

D

D

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

\(a,ĐK:x>0;x\ne4\\ E=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}-2}{2\sqrt{x}}\\ b,x=19-8\sqrt{3}=\left(4-\sqrt{3}\right)^2\\ \Leftrightarrow E=\dfrac{4-\sqrt{3}-2}{2\left(4-\sqrt{3}\right)}=\dfrac{\left(2-\sqrt{3}\right)\left(4+\sqrt{3}\right)}{26}=\dfrac{5-2\sqrt{3}}{26}\\ c,E=-1\Leftrightarrow\sqrt{x}-2=-2\sqrt{x}\\ \Leftrightarrow3\sqrt{x}=2\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(tm\right)\\ d,E=\dfrac{1}{\sqrt{x}}\Leftrightarrow\dfrac{\sqrt{x}-2}{2}=1\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

\(e,E>0\Leftrightarrow\sqrt{x}-2>0\left(2\sqrt{x}>0\right)\Leftrightarrow x>4\\ f,E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}=\dfrac{1}{2}-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\left(-\dfrac{1}{\sqrt{x}}< 0\right)\\ g,\dfrac{1}{E}=\dfrac{2\sqrt{x}}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(4\right)=\left\{-1;0;1;2;4\right\}\left(\sqrt{x}-2>-2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{1;2;3;4;6\right\}\\ \Leftrightarrow x\in\left\{1;9;16;36\right\}\left(x\ne4\right)\\ h,x>4\Leftrightarrow\sqrt{x}-2>0\\ \Leftrightarrow E=\dfrac{\sqrt{x}-2}{2\sqrt{x}}>0\Leftrightarrow E\ge\sqrt{E}\)

a) Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{2x-2}{\sqrt{x}-1}\)

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

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

\(=x+1\)

b) Gọi vận tốc lúc đi của người đó là x(km/h)(Điều kiện: x>0)

Vận tốc lúc về của người đó là: x+5(km/h)

Thời gian ô tô đi từ A đến B là: \(\dfrac{90}{x}\)(h)

Thời gian ô tô đi B về A là: \(\dfrac{90}{x+5}\left(h\right)\)

Theo đề, ta có phương trình:

\(\dfrac{90}{x}-\dfrac{90}{x+5}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{450\left(x+5\right)}{5x\left(x+5\right)}-\dfrac{450x}{5x\left(x+5\right)}=\dfrac{x\left(x+5\right)}{5x\left(x+5\right)}\)

Suy ra: \(x^2+5x-2250=0\)

\(\text{Δ}=5^2-4\cdot1\cdot\left(-2250\right)=9025\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-5-95}{2}=\dfrac{-100}{2}=-50\left(loại\right)\\x_2=\dfrac{-5+95}{2}=\dfrac{90}{2}=45\left(nhận\right)\end{matrix}\right.\)

Vậy: Vận tốc lúc đi là 45km/h

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

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

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

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

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

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

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

b) Mà với \(x\ge0\) và \(x\ne1\) thì 

\(x+\sqrt{x}+1\ge0\) và \(2>0\) nên \(P>0\)

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

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

b: x+căn x+1+1>=1>0

2>0

=>P>0 với mọi x thỏa mãn x>=0 và x<>1

\(=\dfrac{1}{y-x}\cdot x^3\cdot\left(x-y\right)=-x^3\)

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

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

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

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

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

\(=\dfrac{2}{x-1}\)