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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
ĐẶT: \(a=\sqrt[3]{\sqrt{2}-1}\)
=> \(a^3=\sqrt{2}-1\)
=> \(x=a-\frac{1}{a}\)
=> \(x^3=a^3-\frac{1}{a^3}-3a+\frac{3}{a}\)
<=> \(x^3=\sqrt{2}-1-\frac{1}{\sqrt{2}-1}-3\left(a-\frac{1}{a}\right)\)
<=> \(x^3=\frac{\left(\sqrt{2}-1\right)^2-1}{\sqrt{2}-1}-3x\)
<=> \(x^3+3x=\frac{3-2\sqrt{2}-1}{\sqrt{2}-1}\)
<=> \(x^3+3x=\frac{2-2\sqrt{2}}{\sqrt{2}-1}\)
<=> \(x^3+3x=\frac{2\left(1-\sqrt{2}\right)}{\sqrt{2}-1}\)
<=> \(x^3+3x=-2\)
<=> \(x^3+3x+2=0\Rightarrow P=0\)
VẬY \(P=0\)
Đặt \(a=\sqrt[3]{\sqrt{2}-1};b=\frac{1}{\sqrt[3]{\sqrt{2}-1}}\Rightarrow\hept{\begin{cases}x=a-b\\ab=1\end{cases}}\)
Xét \(x^3=\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\)
\(x^3=\left(\sqrt{2}-1\right)-\frac{1}{\sqrt{2}-1}-3x\)
\(\Leftrightarrow x^3=-2-3x\Leftrightarrow x^3+3x+2=0\)
Vậy P=0