Đáp án:

$\begin{array}{l}
B1)\\
a)Dkxd:x\# 0;x\# 1\\
P = \dfrac{{{x^2} + x}}{{{x^2} - 2x + 1}}:\left( {\dfrac{{x + 1}}{x} - \dfrac{1}{{1 - x}} + \dfrac{{2 - {x^2}}}{{{x^2} - x}}} \right)\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}:\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) + x + 2 - {x^2}}}{{x\left( {x - 1} \right)}}\\
 = \dfrac{{x\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}.\dfrac{{x\left( {x - 1} \right)}}{{{x^2} - 1 + x + 2 - {x^2}}}\\
 = \dfrac{{x\left( {x + 1} \right)}}{{x - 1}}.\dfrac{x}{{x + 1}}\\
 = \dfrac{{{x^2}}}{{x - 1}}\\
b)P > 1\\
 \Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} - 1 > 0\\
 \Leftrightarrow \dfrac{{{x^2} - x + 1}}{{x - 1}} > 0\\
 \Leftrightarrow x - 1 > 0\left( {do:{x^2} - x + 1 > 0} \right)\\
 \Leftrightarrow x > 1\\
Vậy\,x > 1\\
c)x > 1\\
P = \dfrac{{{x^2}}}{{x - 1}} = \dfrac{{{x^2} - 1 + 1}}{{x - 1}}\\
 = \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} + \dfrac{1}{{x - 1}}\\
 = x + 1 + \dfrac{1}{{x - 1}}\\
 = x - 1 + \dfrac{1}{{x - 1}} + 2\\
Do:x > 1 \Leftrightarrow x - 1 > 0\\
Theo\,Co - si:\left( {x - 1} \right) + \dfrac{1}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{1}{{x - 1}}}  = 2\\
 \Leftrightarrow x - 1 + \dfrac{1}{{x - 1}} + 2 \ge 2 + 2\\
 \Leftrightarrow P \ge 4\\
 \Leftrightarrow GTNN:P = 4\,Khi:x = 2\\
B2)a){x^2} + \left( {x + 5} \right)\left( {3x - 7} \right) = 25 + 2{x^2}\\
 \Leftrightarrow {x^2} + 3{x^2} - 7x + 15x - 35 = 25 + 2{x^2}\\
 \Leftrightarrow 2{x^2} + 8x - 60 = 0\\
 \Leftrightarrow {x^2} + 4x - 30 = 0\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = 34\\
 \Leftrightarrow x =  - 2 \pm \sqrt {34} \\
Vậy\,x =  - 2 \pm \sqrt {34} \\
b){x^4} - 8{x^3} + 11{x^2} - 8x + 1 = 0\\
 \Leftrightarrow {x^2} - 8x + 11 - \dfrac{8}{x} + \dfrac{1}{{{x^2}}} = 0\\
 \Leftrightarrow {x^2} + \dfrac{1}{{{x^2}}} + 2 - 8.\left( {x + \dfrac{1}{x}} \right) + 9 = 0\\
 \Leftrightarrow {\left( {x + \dfrac{1}{x}} \right)^2} - 8.\left( {x + \dfrac{1}{x}} \right) + 9 = 0\\
Đặt:x + \dfrac{1}{x} = a\\
 \Leftrightarrow {a^2} - 8a + 9 = 0\\
 \Leftrightarrow {\left( {a - 4} \right)^2} = 7\\
 \Leftrightarrow a = 4 \pm \sqrt 7 \\
 \Leftrightarrow x + \dfrac{1}{x} = 4 \pm \sqrt 7 
\end{array}$