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Ta có \(x=\sqrt{\dfrac{2-\sqrt{3}}{2}}=\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{4}}=\sqrt{\dfrac{4-2\sqrt{3}}{4}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{\sqrt{4}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\dfrac{\sqrt{3}-1}{2}\)Ta lại có \(2x^2+2x-1=2.\left(\dfrac{\sqrt{3}-1}{2}\right)^2+2\left(\dfrac{\sqrt{3}-1}{2}\right)-1=2\left(\dfrac{3-2\sqrt{3}+1}{4}\right)+\dfrac{2\left(\sqrt{3}-1\right)}{2}-1=\dfrac{4-2\sqrt{3}}{2}+\sqrt{3}-1-1=\dfrac{2\left(2-\sqrt{3}\right)}{2}+\sqrt{3}-2=2-\sqrt{3}+\sqrt{3}-2=0\)(1)
⇒\(x^3\left(2x^2+2x-1\right)=0\Rightarrow2x^5+2x^4-x^3=0\Rightarrow2x^5+2x^4-x^3-1=-1\Rightarrow\left(2x^5+2x^4-x^3-1\right)^{2016}=\left(-1\right)^{2016}=1\)(2)
Từ (1)⇒\(x\left(2x^2+2x-1\right)=0\Rightarrow2x^3+2x^2-x=0\Rightarrow2x^3+2x^2-x-3=-3\Rightarrow\left(2x^3+2x^2-x-3\right)^{2017}=\left(-3\right)^{2017}\)(3)
Từ (1)⇒\(x^2\left(2x^2+2x-1\right)=0\Rightarrow2x^4+2x^3-x^2=0\Rightarrow2x^4+2x^3-x^2-3=-3\)(4)
Từ (2),(3),(4)⇒\(\left(2x^5+2x^4-x^3-1\right)^{2016}+\dfrac{\left(2x^3+2x^2-x-3\right)^{2017}}{2x^4+2x^3-x^2-3}=1+\dfrac{\left(-3\right)^{2017}}{-3}=1+\left(-3\right)^{2016}=3^{2016}+1\Rightarrow P=3^{2016}+1\)
Ta có \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}=\sqrt{\dfrac{2\sqrt{3}+2}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}-\dfrac{3\left(2\sqrt{3}-2\right)}{\left(2\sqrt{3}-2\right)\left(2\sqrt{3}+2\right)}}=\sqrt{\dfrac{2\left(\sqrt{3}+1\right)}{12-4}-\dfrac{2\left(3\sqrt{3}-3\right)}{12-4}}=\sqrt{\dfrac{\sqrt{3}+1}{4}-\dfrac{3\sqrt{3}-3}{4}}=\sqrt{\dfrac{\sqrt{3}+1-3\sqrt{3}+3}{4}}=\sqrt{\dfrac{4-2\sqrt{3}}{4}}=\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{4}}=\dfrac{\sqrt{3-2\sqrt{3}+1}}{2}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\dfrac{\left|\sqrt{3}-1\right|}{2}=\dfrac{\sqrt{3}-1}{2}\Leftrightarrow2x=\sqrt{3}-1\Leftrightarrow2x+1=\sqrt{3}\Leftrightarrow\left(2x+1\right)^2=3\Leftrightarrow4x^2+4x-2=0\Leftrightarrow2x^2+2x-1=0\)
Ta lại có \(P=\dfrac{4\left(x+1\right)x^{2018}-2x^{2017}+2x+1}{2x^2+3x}=\dfrac{2x^{2017}\left[2\left(x+1\right)x-1\right]+\sqrt{3}}{2x^2+2x-1+x+1}=\dfrac{2x^{2017}\left[2x^2+2x-1\right]+\sqrt{3}}{x+1}=\dfrac{\sqrt{3}}{x+1}=\sqrt{3}:\left(x+1\right)=\sqrt{3}:\left(\dfrac{\sqrt{3}-1}{2}+1\right)=\sqrt{3}:\dfrac{\sqrt{3}+1}{2}=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\left(3-\sqrt{3}\right)}{2}=3-\sqrt{3}\)Vậy khi \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\) thì P=\(3-\sqrt{3}\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình