Giúp mình với ạaaaaa

Giải thích các bước giải:

a.Ta có:

$A+B=(2x+3y)+(2x-y)=2x+3y+2x-y=4x+2y$

$A-B=(2x+3y)-(2x-y)=2x+3y-2x+y=4y$

b.Ta có:

$A+B=(x^2y+x^3-xy^2+2)+(x^3+xy^2-x^2y-7)=x^2y+x^3-xy^2+2+x^3+xy^2-x^2y-7=x^3+x^3+x^2y-x^2y-xy^2+xy^2+2-7=2x^3-5$

$A-B=(x^2y+x^3-xy^2+2)-(x^3+xy^2-x^2y-7)=x^2y+x^3-xy^2+2-x^3-xy^2+x^2y+7=2x^2y-2xy^2+9$

c.Ta có:

$A+B=(2x^2-yz-z^2+1)+(4yz+3x^2+z^2-2)$

$\to A+B=2x^2-yz-z^2+1+4yz+3x^2+z^2-2$

$\to A+B=5x^2+3yz-1$

Ta có:

$A-B=(2x^2-yz-z^2+1)-(4yz+3x^2+z^2-2)$

$\to A-B=2x^2-yz-z^2+1-4yz-3x^2-z^2+2$

$\to A-B=-x^2-5yz-2z^2+3$

d.Ta có:

$A+B=(x^2y+\dfrac32xy^3-\dfrac{11}2x^3y^2+x^3)+(\dfrac12xy^3-x^2y+\dfrac92x^3y^2)$

$\to A+B=x^2y+\dfrac{3}{2}xy^3-\dfrac{11}{2}x^3y^2+x^3+\dfrac{1}{2}xy^3-x^2y+\dfrac{9}{2}x^3y^2$

$\to A+B=x^3+2xy^3-x^3y^2$

Ta có:

$A-B=(x^2y+\dfrac32xy^3-\dfrac{11}2x^3y^2+x^3)-(\dfrac12xy^3-x^2y+\dfrac92x^3y^2)$

$\to A-B=x^2y+\dfrac{3xy^3}{2}-\dfrac{11x^3y^2}{2}+x^3-\dfrac{1}{2}xy^3+x^2y-\dfrac{9}{2}x^3y^2$

$\to A-B=x^3+2x^3y+xy^3-10x^3y^2$