Đáp án:

$\begin{array}{l}
1)A = {x^3} + {y^3} + {z^3} - 3xyz\\
 = {\left( {x + y} \right)^3} - 3{x^2}y - 3x{y^2} + {z^3} - 3xyz\\
 = {\left( {x + y} \right)^3} + {z^3} - 3xy\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right)\left( {{{\left( {x + y} \right)}^2} - \left( {x + y} \right).z + {z^2}} \right)\\
 - 3xyz\left( {x + y + z} \right)\\
 = \left( {x + y + z} \right)\left( {{x^2} + 2xy + {y^2} - xz - yz + {z^2} - 3xy} \right)\\
 = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - xz} \right)\\
2)a)Dkxd:x\# 0;x\# 1\\
P = \dfrac{{{x^2} + x}}{{{x^2} - 2x + 1}}:\left( {\dfrac{{x + 1}}{x} + \dfrac{1}{{x - 1}} + \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 =  - \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{{x^2}}}{{x - 1}} =  - \dfrac{1}{2}\\
 \Leftrightarrow 2{x^2} = 1 - x\\
 \Leftrightarrow 2{x^2} + x - 1 = 0\\
 \Leftrightarrow 2{x^2} + 2x - x - 1 = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {2x - 1} \right) = 0\\
 \Leftrightarrow x =  - 1;x = \dfrac{1}{2}\\
Vậy\,x =  - 1;x = \dfrac{1}{2}\\
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) + 1}}{{x - 1}}\\
 = x + 1 + \dfrac{1}{{x - 1}}\\
 = x - 1 + \dfrac{1}{{x - 1}} + 2\\
Theo\,Co - si:\\
\left( {x - 1} \right) + \dfrac{1}{{x - 1}} \ge 2\sqrt {\left( {x - 1} \right).\dfrac{1}{{x - 1}}}  = 2\\
 \Leftrightarrow \left( {x - 1} \right) + \dfrac{1}{{x - 1}} + 2 \ge 4\\
 \Leftrightarrow P \ge 4\\
 \Leftrightarrow GTNN:P = 4\\
Khi:\left( {x - 1} \right) = \dfrac{1}{{x - 1}}\\
 \Leftrightarrow {\left( {x - 1} \right)^2} = 1\\
 \Leftrightarrow x = 2\left( {tmdk} \right)\\
Vậy\,GTNN:P = 4\,khi:x = 2
\end{array}$

Đáp án+Giải thích các bước giải:

`1)A=x^3+y^3+z^3-3xyz`

`A=x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz`

`A=(x+y)^3+z^3-3xy(x+y+z)`

`A=(x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)`

`A=(x+y+z)[x^2+2xy+y^2-xz-yz+z^2]-3xy(x+y+z)`

`A=(x+y+z)(x^2+y^2+z^2+2xy-3xy-xz-yz)`

`A=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)`