/ kí hiệu là trên

cmr là cái j

Lê Thanh Thùy Ngân 

cmr là chứng minh rằng bạn nhé 

\(A=\dfrac{x+\sqrt{x}+10+\sqrt{x}+3}{x-9}=\dfrac{x+2\sqrt{x}+13}{x-9}\)

Để A>B thì A-B>0

=>\(\dfrac{x+2\sqrt{x}+13}{x-9}-\sqrt{x}-1>0\)

=>\(\dfrac{x+2\sqrt{x}+13-\left(x-9\right)\left(\sqrt{x}+1\right)}{x-9}>0\)

=>\(\dfrac{x+2\sqrt{x}+13-x\sqrt{x}-x+9\sqrt{x}+9}{x-9}>0\)

=>\(\dfrac{-x\sqrt{x}+11\sqrt{x}+22}{x-9}>0\)

TH1: \(\left\{{}\begin{matrix}-x\sqrt{x}+11\sqrt{x}+22>0\\x-9>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}< 4.05\\x>9\end{matrix}\right.\Leftrightarrow9< x< 16.4025\)

TH2: \(\left\{{}\begin{matrix}-x\sqrt{x}+11\sqrt{x}+22< 0\\x-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>4.05\\0< x< 9\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

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

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

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

b: Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1-2⋮\sqrt{x}+1\)

=>căn x+1 thuộc {1;2}

=>căn x thuộc {0;1}

mà x<>1

nên x=0