\(x^5+x-1\)

\(=x^5+x^2-\left(x^2-x+1\right)\)

\(=x^2.\left(x^3+1\right)-\left(x^2-x+1\right)\)

\(=x^2.\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)

\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)

 ab + ac + b² + c² + 2bc

= a(b + c) + b² + 2bc + c²

= a(b + c) + (b + c)²

= (b + c)(a + b + c)

2) 

\(A=2x^2+2x+y^2-2xy=x^2-2xy+y^2+x^2+2x+1-1\)

\(=\left(x-y\right)^2+\left(x+1\right)^2-1\ge-1\)

Dấu \(=\)khi \(\hept{\begin{cases}x-y=0\\x+1=0\end{cases}}\Leftrightarrow x=y=-1\).

Vậy GTNN của \(A\)là \(-1\)đạt tại \(x=y=-1\).

\(B=2a^2+b^2+c^2-ab+ac+bc\)

\(2B=4a^2+2b^2+2c^2-2ab+2ac+2bc\)

\(=a^2-2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2+2a^2\)

\(=\left(a-b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2+2a^2\ge0\)

Dấu \(=\)khi \(a=b=c=0\).

Vậy GTNN của \(B\)là \(0\)đạt tại \(a=b=c=0\).

1. 

a) \(2x^2+2x+1=x^2+x^2+2x+1=x^2+\left(x+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)(vô nghiệm) 

suy ra đpcm

b) \(x^2+y^2+2xy+2y+2x+2=\left(x+y\right)^2+2\left(x+y\right)+1+1=\left(x+y+1\right)^2+1>0\)

c) \(3x^2-2x+1+y^2-2xy+1=x^2-2xy+y^2+x^2-2x+1+x^2+1\)

\(=\left(x-y\right)^2+\left(x-1\right)^2+x^2+1>0\)

d) \(3x^2+y^2+10x-2xy+26=x^2-2xy+y^2+x^2+10x+25+x^2+1\)

\(=\left(x-y\right)^2+\left(x+5\right)^2+x^2+1>0\)