\(B=\left(x^2-8x\right)\left(x^2-8x+24\right)\)
Đặt \(x^2-8x+12=t\) ta có:
\(B=\left(t-12\right)\left(t+12\right)=t^2-144\ge-144\)
Dấu "=" xảy ra khi \(t^2=0\Leftrightarrow t=0\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\Leftrightarrow x=2;x=6\)
\(C=5x^2+9y^2-6xy-12x+13\)
\(=\left(x^2-6xy+9y^2\right)+\left(4x^2-12x+9\right)+4\)
\(=\left(x-3y\right)^2+\left(2x-3\right)^2+4\ge4\)
Dấu "=" xảy ra tại \(x=\frac{3}{2};y=\frac{1}{2}\)
Câu D nhân 4 vào cho nó đẹp:v
\(4D=4x^2+4y^2+4xy-48x-48y+400\)
\(=\left(4x^2+2.2x.y+y^2\right)-24\left(2x+y\right)+3y^2-24y+400\)
\(=\left(2x+y\right)^2-2.\left(2x+y\right).12+12^2+3y^2-24y+256\)
\(=\left(2x+y-12\right)^2+3\left(y-4\right)^2+208\ge208\)
Suy ra \( D\ge\frac{208}{4}=52\)
Đẳng thức xảy ra khi x = y = 4
