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a. \(Z=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\)\(\left(x\ge0,x\ne1\right)\)
\(Z=\dfrac{x\sqrt{x}-\sqrt{x}-x\sqrt{x}+x-2\sqrt{x}+2}{x-1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}\)
\(Z=\dfrac{-3\sqrt{x}+x+2}{x-1}:\dfrac{x-4}{x-1}=\dfrac{x-3\sqrt{x}+2}{x-1}.\dfrac{x-1}{x-4}\)
\(Z=\dfrac{x-3\sqrt{x}+2}{x-4}\)
b. \(Z=\dfrac{x-3\sqrt{x}+2}{x-4}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{x-4}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2x-6\sqrt{x}+4-x+4}{2x-8}< 0\)
\(\Leftrightarrow\dfrac{x-6\sqrt{x}+8}{2x-8}< 0\)\(\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{2\left(x-4\right)}< 0\)
*\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)< 0\\2\left(x-4\right)>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-2< 0\&\sqrt{x}-4>0\\\sqrt{x}-2>0\&\sqrt{x}-4< 0\end{matrix}\right.\\2\left(x-4\right)>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 4\&x>16\left(l\right)\\16>x>4\end{matrix}\right.\\x>4\end{matrix}\right.\)
* \(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)>0\\2\left(x-4\right)< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-2>0\&\sqrt{x}-4>0\\\sqrt{x}-2< 0\&\sqrt{x}-4< 0\end{matrix}\right.\\2\left(x-4\right)< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>16\\x< 4\end{matrix}\right.\\x< 4\end{matrix}\right.\) \(\Leftrightarrow16>x>4\)
Vậy: \(Z< \dfrac{1}{2}\Leftrightarrow16>x>4\)
1 , ĐKXĐ : \(x\ge0,x\ne1\)
Với điều kiện xác định trên phương trình đã cho thánh :
\(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1-2\left(\sqrt{x}+1\right)+x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
b: \(B=\left(2-\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\cdot\left(2-\dfrac{\sqrt{a}\left(5-\sqrt{b}\right)}{-\left(5-\sqrt{b}\right)}\right)\)
\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=4-a\)
c: \(C=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+2\right)\left(2-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)\)
=4-x
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
1: Sửa đề: \(B=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
2: Để B<=-1/2 thì B+1/2<=0
=>-3/căn x+3+1/2<=0
=>-6+căn x+3<=0
=>căn x<=3
=>0<x<9
3: Để B là số nguyên thì \(\sqrt{x}+3=3\)
=>x=0
b: \(A=\dfrac{P}{Q}=\dfrac{x+3}{\sqrt{x}-2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{x+3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}\ge2\sqrt{3}\)
Dấu '=' xảy ra khi \(x=3\)
a) ĐKXĐ: \(x>0;x\ne1;x\ne4\), rút gọn: \(Y=\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)+1}{\sqrt{x}-1}:\dfrac{\left(\sqrt{x}-1\right)^2-1}{\sqrt{x}-1}=\dfrac{x-\sqrt{x}-3\sqrt{x}+3+1}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{x-2\sqrt{x}+1-1}=\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(a.Y=\left(\sqrt{x}-3+\dfrac{1}{\sqrt{x}-1}\right):\left(\sqrt{x}-1+\dfrac{1}{1-\sqrt{x}}\right)=\dfrac{x-4\sqrt{x}+3+1}{\sqrt{x}-1}:\dfrac{x-2\sqrt{x}+1-1}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\left(x>0;x\ne1;x\ne4\right)\)
\(b.Y< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-4}{2\sqrt{x}}< 0\)
Do : \(2\sqrt{x}>0\)\(\Leftrightarrow\sqrt{x}-4< 0\Leftrightarrow x< 16\)
Kết hợp với ĐKXĐ : \(0< x< 16\left(x\ne1;x\ne4\right)\)