Câu 2.

Kết hợp với giả thiết nêu ra ở đề bài, ta có vài biến đổi sau: 

\(\dfrac{x}{y^3-1}=\dfrac{x}{\left(y-1\right)\left(y^2+y+1\right)}=\dfrac{x}{\left[y-\left(x+y\right)\right]\left(y^2+y+1\right)}=-\dfrac{1}{y^2+y+1}\)  \(\left(1\right)\)

\(\dfrac{y}{x^3-1}=\dfrac{y}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{y}{\left[x-\left(x+y\right)\right]\left(x^2+x+1\right)}=-\dfrac{1}{x^2+x+1}\)  \(\left(2\right)\)

Mặt khác, ta lại có: \(\left(x^2+x+1\right)\left(y^2+y+1\right)=x^2y^2+xy^2+y^2+x^2y+xy+y+x^2+x+1\)

\(=x^2y^2+\left[x^2+xy\left(x+y\right)+xy+y^2\right]+\left(x+y\right)+1=x^2y^2+\left(x+y\right)^2+2=x^2y^2+3\)

Khi đó,  trừ đẳng thức  \(\left(1\right)\)  cho  đẳng thức  \(\left(2\right)\)  vế theo vế, ta được:

\(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}=\dfrac{1}{x^2+x+1}-\dfrac{1}{y^2+y+1}=\dfrac{\left(y-x\right)\left(x+y+1\right)}{\left(x^2+x+1\right)\left(y^2+y+1\right)}=\dfrac{-2\left(x-y\right)}{x^2y^2+3}\)

Vậy,  \(\dfrac{x}{y^3-1}-\dfrac{y}{x^3-1}+\dfrac{2\left(x-y\right)}{x^2y^2+3}=-\dfrac{2\left(x-y\right)}{x^2y^2+3}+\dfrac{2\left(x-y\right)}{x^2y^2+3}=0\)

Câu 3.

$ {a^2} - 4a + 1 = 0\\ \Leftrightarrow {\left( {a - 2} \right)^2} - 3 = 0\\ \Leftrightarrow {\left( {a - 2} \right)^2} - {\left( {\sqrt 3 } \right)^2} = 0\\ \Leftrightarrow \left( {a - 2 + \sqrt 3 } \right)\left( {a - 2 - \sqrt 3 } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} a - 2 + \sqrt 3 = 0\\ a - 2 - \sqrt 3 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} a = 2 - \sqrt 3 \\ a = 2 + \sqrt 3 \end{array} \right.\\ *a = 2 - \sqrt 3 \Rightarrow P = \dfrac{{{{\left( {2 - \sqrt 3 } \right)}^4} + {{\left( {2 - \sqrt 3 } \right)}^2} + 1}}{{{{\left( {2 - \sqrt 3 } \right)}^2}}}\\ = \dfrac{{97 - 56\sqrt 3 + 4 - 4\sqrt 3 + 3 + 1}}{{4 - 4\sqrt 3 + 3}} = \dfrac{{105 - 60\sqrt 3 }}{{7 - 4\sqrt 3 }} = \dfrac{{15\left( {7 - 4\sqrt 3 } \right)}}{{7 - 4\sqrt 3 }} = 15\\ *a = 2 + \sqrt 3 \Rightarrow P = \dfrac{{{{\left( {2 + \sqrt 3 } \right)}^4} + {{\left( {2 + \sqrt 3 } \right)}^2} + 1}}{{{{\left( {2 + \sqrt 3 } \right)}^2}}}\\ = \dfrac{{97 + 56\sqrt 3 + 4 + 4\sqrt 3 + 3 + 1}}{{4 + 4\sqrt 3 + 3}} = \dfrac{{105 + 60\sqrt 3 }}{{7 + 4\sqrt 3 }}\\ = \dfrac{{15\left( {7 + 4\sqrt 3 } \right)}}{{7 + 4\sqrt 3 }} = 15 $