\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)

\(=\left(x^2+4x+4-3\right)^2-12\left(x+2\right)^2+2093\)

\(=\left[\left(x+2\right)^2-3\right]^2-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-6\left(x+2\right)^2+9-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+2102\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+81+2021\)

\(=\left[\left(x+2\right)^4-18\left(x+2\right)^2+81\right]+2021\)

\(=\left[\left(x+2\right)^2-9\right]^2+2021\)

\(=\left[\left(x+2-3\right)\left(x+2+3\right)\right]^2+2021\)

\(=\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\)

Vì \(\left[\left(x-1\right)\left(x+5\right)\right]^2\ge0\forall x\)

\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\ge2021\)\(\forall x\)

hay \(P\ge2021\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(minP=2021\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Sai đề nha.

\(0^n+0^n=0^n\)

n={1;2;3}

6a( x - 3y ) - 8b( 3y - x )

= 6a( x - 3y ) + 8b( x - 3y )

= 2( x - 3y )( 3a + 4b )

\(6a\left(x-3y\right)-8b\left(3y-x\right)\)

\(=6a\left(x-3y\right)+8b\left(x-3y\right)\)

\(=\left(6a+8b\right)\left(x-3y\right)=2\left(3a+4b\right)\left(x-3y\right)\)