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a) Ta có: \(3x+2\sqrt{3x}+4=\left(\sqrt{3x}+1\right)^2+3>0;1+\sqrt{3x}>0,\forall x\ge0\), nên đk để A có nghĩa là
\(\left(\sqrt{3x}\right)^3-8-\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)\ne0;x\ge0\Leftrightarrow\sqrt{3x}\ne2\Leftrightarrow0\le x\ne\frac{4}{3}\)
A=\(\left(\frac{6x+4}{\left(\sqrt{3x}\right)^3-2^3}-\frac{\sqrt{3x}}{3x+2\sqrt{3x}+4}\right)\left(\frac{1+\left(\sqrt{3x}\right)^3}{1+\sqrt{3x}}-\sqrt{3x}\right)\)
\(=\left(\frac{6x+4-\left(\sqrt{3x}-2\right)\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-\sqrt{3x}+1-\sqrt{3x}\right)\)
\(=\left(\frac{3x+4+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\left(3x-2\sqrt{3x}+1\right)\)
\(=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}\left(0\le x\ne\frac{4}{3}\right)\)
b) \(A=\frac{\left(\sqrt{3x}-1\right)^2}{\sqrt{3x}-2}=\frac{\left(\sqrt{3x}-2\right)^2+2\left(\sqrt{3x}-2\right)+1}{\sqrt{3x}-2}=\sqrt{3x}+\frac{1}{\sqrt{3x}-2}\)
Với \(x\ge0\), để A là số nguyên thì \(\sqrt{3x}-2=\pm1\Leftrightarrow\orbr{\begin{cases}\sqrt{3x}=3\\\sqrt{3x}=1\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=9\\3x=1\end{cases}\Leftrightarrow}x=3}\) (vì \(x\in Z;x\ge0\))
Khi đó A=4
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)
\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}< 0\)
\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)
\(\Leftrightarrow\sqrt{x}-6< 0\)
\(\Leftrightarrow x< 36\)
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)
a: \(A=\left(\dfrac{6x+4-3x+2\sqrt{3x}}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\cdot\left(3x-\sqrt{3x}+1-\sqrt{3x}\right)\)
\(=\left(\dfrac{3x+2\sqrt{3x}+4}{\left(\sqrt{3x}-2\right)\left(3x+2\sqrt{3x}+4\right)}\right)\cdot\left(3x-2\sqrt{3x}+1\right)\)
\(=\sqrt{3x}-1\)
b: Để A là số nguyên thì căn 3x là số nguyên
=>3x=k2(k∈N)
hay \(x=\dfrac{k^2}{3}\)
