1) \(A=x\left(x-6\right)+10=x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=3\)
\(B=x^2-2x+9y^2-6y+3\)
\(B=\left(x^2-2x+1\right)+\left(9y^2-6y+1\right)+1\)
\(B=\left(x-1\right)^2+\left(3y-1\right)^2+1\ge1>0\)
Dấu "=" xảy ra khi: \(x=y=1\)
2) \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3\ge-3\)
Dấu "=" xảy ra khi: \(x=2\)
\(B=4x^2+4x+11=4x^2+4x+1+10=\left(2x+1\right)^2+10\ge10\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{2}\)
\(C\) mk nghĩ đề sai
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)\)
\(C=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)\)
\(C=\left(x^2+5x+4\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x+5-1\right)\left(x^2+5x+5+1\right)\)
\(C=\left(x^2+5x+5\right)^2-1\)
\(C=\left(x^2+5x+\dfrac{25}{4}-\dfrac{5}{4}\right)^2-1\)
\(C=\left[\left(x+\dfrac{5}{2}\right)^2-\dfrac{5}{4}\right]^2-1\ge\dfrac{9}{16}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{5}{2}\)
\(D=4x-x^2+1=-\left(x^2-4x-1\right)=-\left(x^2-4x+4-5\right)=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)
Dấu "=" xảy ra khi: \(x=2\)
\(E=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)=-\left(x+4\right)^2+21\le21\)
Dấu "=" xảy ra khi: \(x=-4\)
