\(x^2-2xy+2y^2+2y+5=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)+4=\left(x-y\right)^2+\left(y+1\right)^2+4\)

Do \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\\\left(y+1\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y\)

\(\Rightarrow\left(x-y\right)^2+\left(y+1\right)^2+4>0\) ; \(\forall x;y\)

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

a : x² + 4x + 7 = (x + 2)² + 3 > 0

b : 4x² - 4x + 5 = (2x - 1)² + 4 > 0

c : x² + 2y² + 2xy - 2y + 3 = (x + y)² + (y - 1)² + 2 > 0

d : 2x² - 4x + 10 = 2(x - 1)² + 8 > 0

e : x² + x + 1 = (x + 0,5)² + 0,75 > 0

f : 2x² - 6x + 5 = 2(x - 1,5)² + 0,5 > 0

\(a,P=5x\left(2-x\right)-\left(x+1\right)\left(x+9\right)\)

\(=10x-5x^2-\left(x^2+x+9x+9\right)\)

\(=10x-5x^2-x^2-x-9x-9\)

\(=\left(10x-x-9x\right)+\left(-5x^2-x^2\right)-9\)

\(=-6x^2-9\)

Ta thấy: \(x^2\ge0\forall x\)

\(\Rightarrow-6x^2\le0\forall x\)

\(\Rightarrow-6x^2-9\le-9< 0\forall x\)

hay \(P\) luôn nhận giá trị âm với mọi giá trị của biến \(x\).

\(b,Q=3x^2+x\left(x-4y\right)-2x\left(6-2y\right)+12x+1\)

\(=3x^2+x^2-4xy-12x+4xy+12x+1\)

\(=\left(3x^2+x^2\right)+\left(-4xy+4xy\right)+\left(-12x+12x\right)+1\)

\(=4x^2+1\)

Ta thấy: \(x^2\ge0\forall x\)

\(\Rightarrow4x^2\ge0\forall x\)

\(\Rightarrow4x^2+1\ge1>0\forall x\)

hay \(Q\) luôn nhận giá trị dương với mọi giá trị của biến \(x\) và \(y\).

#\(Toru\)

\(A=-x^2+3x-7\)

\(=-\left(x^2-3x+7\right)\)

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{19}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}< 0\forall x\)

\(3x-7-x^2=-\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{19}{4}=-\left(x-\dfrac{3}{2}\right)^2-\dfrac{19}{4}\le-\dfrac{19}{4}< 0\)

\(=\dfrac{3x\left(-x^2\right)}{3x}+\dfrac{2}{3x}-\dfrac{3x}{3x}=\dfrac{-3x^3+2-3x}{3x}\)

\(=\dfrac{-x^2+2-3x}{1}=-\left(x^2-2+3x\right)\)

vậy bt A luôn......

a: \(P\left(x\right)=2x^3-x^4+1+2x^2+5x^4-x^3\)

\(=2x^3-x^3+\left(5x^4-x^4\right)+2x^2+1\)

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

\(Q\left(x\right)=-3x^4-1+5x^3-x^2-6x^2-4x^3\)

\(=-3x^4+\left(5x^3-4x^3\right)+\left(-x^2-6x^2\right)-1\)

\(=-3x^4+x^3-7x^2-1\)

b: \(P\left(-2\right)=4\cdot\left(-2\right)^4+\left(-2\right)^3+2\cdot\left(-2\right)^2+1\)

\(=4\cdot16-8+8+1\)

=64+1

=65

c: P(x)+Q(x)

\(=4x^4+x^3+2x^2+1-3x^4+x^3-7x^2-1=x^4+2x^3-5x^2\)

d: Q(x)-P(x)

\(=-3x^4+x^3-7x^2-1-4x^4-x^3-2x^2-1=-7x^4-9x^2-2\le-2<0\forall x\)

a: \(P\left(x\right)=2x^3-x^4+1+2x^2+5x^4-x^3\)

\(=2x^3-x^3+\left(5x^4-x^4\right)+2x^2+1\)

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

\(Q\left(x\right)=-3x^4-1+5x^3-x^2-6x^2-4x^3\)

\(=-3x^4+\left(5x^3-4x^3\right)+\left(-x^2-6x^2\right)-1\)

\(=-3x^4+x^3-7x^2-1\)

b: \(P\left(-2\right)=4\cdot\left(-2\right)^4+\left(-2\right)^3+2\cdot\left(-2\right)^2+1\)

\(=4\cdot16-8+8+1\)

=64+1

=65

c: P(x)+Q(x)

\(=4x^4+x^3+2x^2+1-3x^4+x^3-7x^2-1=x^4+2x^3-5x^2\)

d: Q(x)-P(x)

\(=-3x^4+x^3-7x^2-1-4x^4-x^3-2x^2-1=-7x^4-9x^2-2\le-2<0\forall x\)

Không biê

\(A=x^2+2y^2-2xy-2y+15\)

\(=\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)+14>14>0\)

Vậy : \(A>0\)