Ta có: P(x)+Q(x)= x²y-2xy²+5xy+3+3xy²+5x²y-7xy+2=6x²y+xy²-2xy+5

• Make P(x) x^2*y – 2*x*y^2 + 5*x*y +3
• Make Q(x) 3*x*y^2 + 5*x^2*y – 7*x*y + 2
• Expand P(x) + Q(x)

P(x)= x^2 - 2xy^2 + 5xy+ 3
Q(x)= 5x^2y + 3xy^2- 7xy+2
P(x)+Q(x)= 6x^2y + xy^2 - 2xy +5

a) (5x²y-5xy²+xy) + (xy-x²y²+5xy²)

= 5x²y-5xy²+xy+xy-x²y²+5xy²

= 5x²y+(5xy²-5xy²)+(xy+xy)-x²y²

= 5x²y+2xy-x²y²

b) (x²+y²+z²) + (x²-y²+z²)

= x²+y²+z²+x²-y²+z²

= (x²+x²)+(y²-y²)+(z²+z²)

= 2x²+2z²

a)( \(5x^2y\)\(-\) \(5xy^2\) \(+\) \(xy\)) + (\(xy\) \(-\) \(x^2y^2\) \(+\) \(5xy^2\))

= \(5x^2y-5xy^2+xy+xy-x^2y^2+5xy^2\)

= \(5x^2y+2xy-x^2y^2\)

b) \(\left(x^2+y^2+z^2\right)+\left(x^2-y^2+z^2\right)\)

= \(x^2+y^2+z^2+x^2-y^2+z^2\)

=\(2x^2+2z^2\)

=\(2\left(x+z\right)^2\)

\(P=x^3+x^2y-5x^2-x^2y-xy^2+5xy+3\left(x+y\right)+2000\\ =x^2\left(x+y-5\right)-xy\left(x+y-5\right)+3\left(x+y-5\right)+2015\\ =x^2\left(5-5\right)-xy\left(5-5\right)+3\left(5-5\right)+2015\\ =2015\)

`P = x^3 + x^2 - 5x^2 - x^2y + xy^2 + 5xy + 3(x+y) + 2000`

`P = x^2(x+y) - (x+y)x^2 - xy(x+y) + (x+y)xy + 3(x+y) + 2000`

`P = 0 + 0 + 3.5 + 2000`

`P = 2015`

B6:

Ta có: \(\hept{\begin{cases}P\left(-1\right)=a-b+c\\P\left(-2\right)=4a-2b+c\end{cases}}\)

=> \(P\left(-1\right)+P\left(-2\right)=5a-3b+2c\)

Mà theo đề bài \(5a-3b+2c=0\)

=> \(P\left(-1\right)+P\left(-2\right)=0\Rightarrow P\left(-1\right)=-P\left(-2\right)\)

Thay vào ta được: \(P\left(-1\right).P\left(-2\right)=-P\left(-2\right).P\left(-2\right)=-P\left(-2\right)^2\le0\left(\forall a,b,c\right)\)

=> đpcm

B5:

Ta có:

P+Q+R

= 5x²y²-xy-2y³-y²+5x⁴-2x²y²-5xy+y³-3y²+2x⁴-x²y²+6xy+y³+6y²+7

= x²y²+2y²+7x⁴+7

Mà \(x^2y^2\ge0;2y^2\ge0;7x^4\ge0\left(\forall x,y\right)\)

=> \(x^2y^2+2y^2+7x^4+7\ge7\)

=> Tổng 3 đa thức P,Q,R luôn dương

=> Trong 3 đa thức đó luôn tồn tại 1 đa thức lớn hơn 0

=> đpcm