3(x⁴ + x² + 1 ) - ( x² + x + 1 )² 

= 3x² ( x² + 1 + 1 ) - ( x² + x + 1 ) ( x² + x + 1 )

= 3x² ( x² + 1 + 1 )- ( x⁴ + x² + 1 )

= 3x² ( x² + 1 + 1 ) - x² ( x² + 1 + 1 )

=  x² + 2 - 2x² 

=  -x² + 2

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

\(=3x^4+3x^2+3-x^4-x^2-1-2x^3-2x^2-2x\)

\(=2x^4-2x^3-2x+2\)

\(=2.\left(x^4-x^3-x+1\right)\)

\(=2.\left[\left(x^4-x^3\right)-\left(x-1\right)\right]\)

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

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

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

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

Tham khảo nhé~

\(x^2-2=5\)

\(\Leftrightarrow x^2=7\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=\sqrt{7}\\x=-\sqrt{7}\end{cases}}\)

\(2x^2-7x+5\)

\(=2x^2-2x-5x+5\)

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

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

Chúc bạn học tốt.

\(\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\Leftrightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}-\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\)

\(\Leftrightarrow\left(\frac{x^2}{a^2+b^2+c^2}-\frac{x^2}{a^2}\right)+\left(\frac{y^2}{a^2+b^2+c^2}-\frac{y^2}{b^2}\right)+\left(\frac{z^2}{a^2+b^2+c^2}-\frac{z^2}{c^2}\right)=0\)

\(\Leftrightarrow x^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)+y^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)+z^2.\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)=0\)

vì \(a,b,c\ne0\Rightarrow\hept{\begin{cases}\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{a^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{b^2}\right)\ne0\\\left(\frac{1}{a^2+b^2+c^2}-\frac{1}{c^2}\right)\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}}\Rightarrow x=y=z=0\Rightarrow P=0+\frac{11}{2011}=\frac{11}{2011}\)

\(Q=\frac{2x+1}{x^2+2}\)

\(Q=\frac{x^2+2-x^2+2x-1}{x^2+2}\)

\(Q=\frac{x^2+2}{x^2+2}-\frac{x^2-2x+1}{x^2+2}\)

\(Q=1-\frac{\left(x-1\right)^2}{x^2+2}\)

Vì \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\frac{\left(x-1\right)^2}{x^2+2}\ge0\forall x\)

\(\Rightarrow Q\le1-0=1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

Vậy \(Q_{max}=1\Leftrightarrow x=1\)