Tính: a) $\frac{1}{7}x^{2}y^{3}(-\frac{14}{3}xy^{2})-\frac{1}{2}xy(x^{2}y^{4})$ b) $\frac{2}{5}x^{4}y^{5}(\frac{5}{4}x^{2}y^{3})+\frac{5}{3}x^{2}y^{3}(-\frac{9}{10}x^{4}y^{5})$

Đáp án:

$a) \dfrac{1}{7}x^2y^3.(-\dfrac{14}{3}xy^2)-\dfrac{1}{2}xy.(x^2y^4)$

$=$ {  $[\dfrac{1}{7}.(-\dfrac{14}{3})].(x^2.x).(y^3.y^2)]$ } $- [(\dfrac{1}{2}.1).(x.x^2).(y.y^4)]$

$= \dfrac{-2}{3}x^3y^5-\dfrac{1}{2}x^3y^5$

$=-\dfrac{7}{6}x^3y^5$

$b) \dfrac{2}{5}x^4y^5.(\dfrac{5}{4}x^2y^3)+\dfrac{5}{3}x^2y^3.(-\dfrac{9}{10}x^4y^5)$

$=[(\dfrac{2}{5}.\dfrac{5}{4}).(x^4.x^2).(y^5.y^3)]+$ { $[\dfrac{5}{3}.(-\dfrac{9}{10})].(x^2.x^4).(y^3.y^5)$ }

$= \dfrac{1}{2}x^6y^8-\dfrac{3}{2}x^6y^8$

$= -x^6y^8$

Giải thích các bước giải:

`a)`

`1/7x^2y^3(-14/3xy^2) - 1/2xy(x^2y^4)`

`= 1/7 . (-14/3) . x^2y^3 . xy^2 - 1/2xy(x^2y^4)`

`= (-2)/3x^3y^5 - 1/2x^3y^5`

`= x^3y^5((-2)/3 - 1/2)`

`= (-7)/6x^3y^5`

`b)`

`2/5x^4y^5(5/4x^2y^3) + 5/3x^2y^3(-9/10x^4y^5)`

`= 1/2x^6y^8 + (-3)/2x^6y^8`

`= x^6y^8(1/2 + (-3)/2)`

`= -x^6y^8`