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a) \(M=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{6\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)-6\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
b) \(M=\dfrac{\sqrt{x}-3}{\sqrt{x}+2}=1-\dfrac{5}{\sqrt{x}+2}\in Z\)
\(\Rightarrow\sqrt{x}+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)
Do \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}\in\left\{3\right\}\Rightarrow x=9\left(tm\right)\)
a) \(M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\left(x>0\right)\)
\(=\left(\dfrac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+4}{\sqrt{x}+1}\)
b) Ta có: \(\sqrt{x}+4>\sqrt{x}+1\Rightarrow\dfrac{\sqrt{x}+4}{\sqrt{x}+1}>1\)
c) \(\dfrac{\sqrt{x}+4}{\sqrt{x}+1}=1+\dfrac{3}{\sqrt{x}+1}\)
Ta có: \(\left\{{}\begin{matrix}3>0\\\sqrt{x}+1>0\end{matrix}\right.\Rightarrow1+\dfrac{3}{\sqrt{x}+1}>1\Rightarrow M>1\)
Lại có: \(\sqrt{x}+1>1\left(x>0\right)\Rightarrow\dfrac{3}{\sqrt{x}+1}< 3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}< 4\Rightarrow M< 4\)
\(\Rightarrow1< M< 4\Rightarrow M\in\left\{2;3\right\}\)
\(M=2\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=2\Rightarrow\dfrac{3}{\sqrt{x}+1}=1\Rightarrow\sqrt{x}+1=3\)
\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(M=3\Rightarrow1+\dfrac{3}{\sqrt{x}+1}=3\Rightarrow\dfrac{3}{\sqrt{x}+1}=2\Rightarrow2\sqrt{x}+2=3\)
\(\Rightarrow2\sqrt{x}=1\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
a. \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right):\dfrac{2\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-\left(x^2\sqrt{x}-x^2+x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-x^2\sqrt{x}+x^2-x+\sqrt{x}}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x^2-2x}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2\left(x^2-x\right)}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=2.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}=\dfrac{x-1}{\left(\sqrt{x}-1\right)^2}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b. \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để A có giá trị nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) \(\Leftrightarrow2⋮\left(\sqrt{x}-1\right)\)\(\Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3;-1\right\}\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{2;0;3\right\}\Leftrightarrow x\in\left\{4;0;9\right\}\)
Vậy để A có giá trị nguyên thì \(x\in\left\{4;0;9\right\}\)
a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\dfrac{-4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)
=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)
=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)
=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)
Để BPT luôn đúng thì m<-0,3
