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

a) \(5x^2-12xy+9y^2-4x+4=\left(4x^2-12xy+9y^2\right)+x^2-4x+4=\left(2x-3y\right)^2+\left(x-2\right)^2\ge0\)
b) \(-x^2-2y^2+12x-4y+7=-\left(x^2-12x+36\right)-2\left(y^2+2y+1\right)+45=-\left(x-6\right)^2-2\left(y+1\right)^2+45\le45\)

c)\(4y^2+10x^2+12xy+6x+7=\left(4y^2+12xy+9x^2\right)+x^2+6x+9-2=\left(2y+3x\right)^2+\left(x+3\right)^2-2\ge-2\)

d) \(3-10x^2-4xy-4y^2=3-\left(4y^2+4xy+x^2\right)-9x^2=-\left(2y+x\right)^2-9x^2+3\le3\)

e)\(x^2-5x+y^2-xy-4y+16=\left(\frac{1}{2}x^2-xy+\frac{1}{2}y^2\right)+\frac{1}{2}\left(x^2-10x+25\right)+\frac{1}{2}\left(y^2-8y+16\right)-\frac{9}{2}=\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x-5\right)^2+\frac{1}{2}\left(y-4\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)Phần e) mới nghĩ đk v, tui biết đáp án sao do k xảy ra dấu bằng

a: \(P+\left(3x^2-4xy\right)=6y^2-9xy+x^2\)

=>\(P=x^2-9xy+6y^2-3x^2+4xy=-2x^2-5xy+6y^2\)

b: \(\left(4y^2-8xy\right)-P=5x^2-12xy+4y^2\)

=>\(P=4y^2-8xy-5x^2+12xy-4y^2=-5x^2+4xy\)

c: \(P-\left(x^2-2y^2+3z^2\right)+\left(3x^2-y^2+2z^2\right)=2x^2-3y^2-z^2\)

=>\(P=2x^2-3y^2-z^2+x^2-2y^2+3z^2-3x^2+y^2-2z^2\)

=>P=\(-4y^2\)

`9x^2+4y^2-12xy+6x-4y+1`

`=(3x)^2-2.3x.2y+(2y)^2+2(3x-2y)+1`

`=(3x-2y)^2+2(3x-2y)+1`

`=(3x-2y+1)^2`

9x²+4y²-12xy+6x-4y+1=(3x-2y+1)²