đặt` √x=a`

`P=((2a^2+1)/(a^3-1)+(a)/(a^2+a+1))×((a^3+1)/(a+1)-a)`

`⇔P=(2a^2+1+a(a-1))/(a^3-1)×(a^2-a+1-a)`

`⇔P=(3a^2-a+1)/(a^3-1)×(a-1)^2`

`⇔P=(3a^2-a+1)/(a^2+a+1) ×(a-1)`

`⇔P=(3x-√x+1)/(x+√x+1) ×(√x-1)`

\begin{cases} 3x^2-2xy=160\\x^2-3xy-2y^2=8 \end{cases}

⇔\begin{cases} 3x^2=2xy+160\\x^2=3xy+2y^2+8 \end{cases}

`⇔2xy+160=3(3xy+2y^2+8)`

`⇔0=7xy+6y^2-136`

`⇔0=x^2+4xy+4y^2-144`

`⇔(x+2y)^2=144`

`⇔`\(\left[ \begin{array}{l}x+2y=12\\x+2y=-12\end{array} \right.\) 

`⇔`\(\left[ \begin{array}{l}x=12-2y\\x=-12-2y\end{array} \right.\) 

với  `x=12-2y`

`⇔3(12-2y)^2-2(12-2y)y=160`

`⇔y=2`

`⇒x=12-2.2=8`

với  `x=-12-2y`

`⇔3(-12-2y)^2-2(-12-2y)y=160`

`⇔y=-2`

`⇒x=12-2.2=-8`

Giải thích các bước giải:

P=($\frac{2x+1}{(√x-1)(x+√x+1)}$+$\frac{√x(√x-1)}{(√x-1)(x+√x+1)}$ ).($\frac{(1+√x)(1-√x+x)}{√x+1}$ -√x)

P=($\frac{2x+1+x-√x}{(√x-1)(x+√x+1)}$).(1-√x+x-√x) P=$\frac{3x-√x+1}{(√x-1)(x+√x+1)}$. 1+x-2√x P=$\frac{(3x-√x+1)(x-2√x+1)}{(√x-1)(x+√x+1)}$ P=$\frac{(3x-√x+1)(√x-1)^{2}}{(√x-1)(x+√x+1)}$ P=$\frac{(3x-√x+1)(√x-1)}{(x+√x+1)}$

2) $\left \{ {{3x^{2}-2xy=160} \atop {x^{2}-3xy-2y^{2}=8}} \right.$

<=>$\left \{ {{3x^{2}-2xy=160} \atop {3x^{2}-9xy-6y^{2}=24}} \right.$

<=>$\left \{ {{3x^{2}=160+2xy} \atop {3x^{2}=24+9xy+6y^{2}}} \right.$

<=>160+2xy=24+9xy+6$y^{2}$

<=>7xy+6$y^{2}$ -136=0

<=>(3xy+2$y^{2}$+8)+4xy+$y^{2}$ -144

<=>$x^{2}$+4xy+$y^{2}$ =144

<=>$(x+2y)^{2}$=144

<=>x+2y=±12

<=>\(\left[ \begin{array}{l}x+2y=12\\x+2y=-12\end{array} \right.\)

<=>\(\left[ \begin{array}{l}x=12−2y\\x=−12−2y\end{array} \right.\)

với x= 12 -2y

ta có: 3($12-2y)^{2}$ (12−2y)y=160

=>y=2

=>x=12-2.2=8

với x =-12-2y

ta có: 3(−12−2y)2−2(−12−2y)y=160

=>y=-2

=>x=-12+2.2=8