Question

giúp em

Answer 1

a) \(...=x^2+2x+1+y^2+6y+9\)

\(=\left(x+1\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrowđpcm\)

b) Sửa lại đề \(+6yz\rightarrow-6yz\)

 \(...=\left(x^2+9y^2+z^2+6xy-2xz-6yz\right)+x^2+2z^2\)

\(=\left(x+3y-z\right)^2+\left(x^2+2z^2\right)>0\)

\(\Rightarrowđpcm\)

c) Sửa lại đề \(-5xy\rightarrow-6xy\)

 \(...=\left(4x^2+y^2+9z^2+4xy-12xz-6yz\right)+4x^2+2z^2\)

\(=\left(2x+y-3z\right)^2+\left(4x^2+z^2\right)>0\)

\(\Rightarrowđpcm\)

d) \(...=\left(4x^2+4y^2+4z^2+8xy-8xz-8yz\right)+\left(x^2-2xy+y^2\right)+z^2\)

\(=\left(2x+2y-2z\right)^2+\left(x-y\right)^2+z^2>0\)

\(\Rightarrowđpcm\)

Answer 2

a; \(x^2\) + y² + 2\(x\) + 6y + 10

= (\(x^2\) + 2\(x\) + 1)  +(y² + 6y + 9) 

= (\(x+1\))² + (y + 3)²

Vì (\(x+1\))² ≥ 0 \(\forall\) \(x\); (y + 3)² ≥ 0 ∀ y 

⇒ (\(x+1\))² + (y + 3)² ≥ 0 ∀ \(x;y\) 

Vậy \(x^2\) + y² + 2\(x\) + 6y  + 10 không âm (đpcm)