\(Q=2x^2-6x=2\left(x^2-3x\right)=2\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)
\(=2\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
Dấu "=" xảy ra \(< =>\left(x-\frac{3}{2}\right)^2=0< =>x=\frac{3}{2}\)
Vậy MInQ=-9/2 khi x=3/2
\(M=x^2+y^2-x+6y+10=x^2+y^2-x+6y+1+9=\left(x^2-x+1\right)+\left(y^2+6y+9\right)\)
\(=\left[\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\right]+\left(y^2+2.y.3+9\right)=\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]+\left(y+3\right)^2=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu "=" xảy ra \(< =>\left(x-\frac{1}{2}\right)^2=0=>x=\frac{1}{2}\)
và \(\left(y+3\right)^2=0=>y=-3\)
Vậy minM=3/4 khi x=1/2 và y=-3
\(P=x^2-2x+1+4=\left(x-1\right)^2+4\ge4\)
Vậy Min của P = 4 khi x-1=0 => x=1
\(Q=2\left(x^2-3x\right)=2\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}\right)=2\left(x-\frac{3}{2}\right)^2+\frac{9}{2}\ge\frac{9}{2}\)
Vay6=65 Min của Q = 9/2 khi x-3/2=0 => x=3/2
\(M=x^2-2.\frac{1}{2}x+\frac{1}{4}+y^2+2.3y+9+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\left(y+3\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy Min của M = 3/4 khi x=1/2 , y=-3