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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1-|1-3x|+|3x-1|^2\)
\(=\left(|3x-1|-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
\(\Rightarrow minA=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)hoặc \(x=\frac{1}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x=2-\sqrt{3}\)
Dễ thấy x là nghiệm của PT \(x^2-4x+1\)
\(H=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2019\\ H=\left(x^2-4x+1\right)\left(x^3+x^2+5\right)+2019\\ H=2019\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có: \(x+\sqrt{3}=2\Leftrightarrow x-2=-\sqrt{3}\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow x^2-4x+1=0\)
\(B=x^5-3x^4-3x^3+6x^2-20x+2021\)
\(B=\left(x^5-4x^4+x^3\right)+\left(x^4-4x^3+x^2\right)+\left(5x^2-20x+5\right)+2016\)
\(B=x^3\left(x^2-4x+1\right)+x^2\left(x^2-4x+1\right)+5\left(x^2-4x+1\right)+2016\)
Thế \(x^2-4x+1=0\)\(\Rightarrow B=2016.\)
Đk: x = \(5+2\sqrt{7}\)> 5
Đặt A = \(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)
A2 = \(\left(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\right)^2\)
A2 = \(3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)
A2 = \(6x-2\sqrt{9x^2-6x+1}\)
A2 = \(6x-2\sqrt{\left(3x-1\right)^2}\) (vì x > \(\frac{1}{3}\))
A2 = \(6x-2\left(3x-1\right)\)
A2 = \(6x-6x+2\)
A2 = 2
=> A = \(\sqrt{2}\)
Vậy ....
Đặt: \(A=\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)
=> \(A^2=3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)
=> \(A^2=6x-2\sqrt{9x^2-6x+1}\)
=> \(A^2=6x-2\sqrt{\left(3x-1\right)^2}\)
Mà: \(x=5+2\sqrt{7}\Rightarrow x>\frac{1}{3}\Rightarrow3x>1\Rightarrow3x-1>0\)
=> \(A^2=6x-2\left(3x-1\right)\)
=> \(A^2=6x-6x+2=2\)
Mà: \(\sqrt{3x+\sqrt{6x-1}}>\sqrt{3x-\sqrt{6x-1}}\Rightarrow A>0\)
=> \(A=\sqrt{2}\)
VẬY \(A=\sqrt{2}\)