poprosiłbym o zadanie 4 ​



Poprosiłbym O Zadanie 4 class=

Odpowiedź :

Magda

Odpowiedź:

[tex]a)\ \ 2x^3\cdot3x=6x^4\\\\b)\ \ \frac{1}{\not3_{1}}xy\cdot\not6^2y^2= xy\cdot2y^2=2xy^3\\\\c)\ \ 6b\cdot(-a)^2\cdot(-\frac{1}{2})a^3b=6ba^2\cdot(-\frac{1}{2})a^3b=\not6^3a^2b\cdot(-\frac{1}{\not2_{1}})a^3b=-3a^2b\cdot a^3b=-3a^5b^2[/tex]

a)

[tex]2 {x}^{3} \times 3x = 6 {x}^{4} [/tex]

b)

[tex] \frac{1}{3} xy \times 6 {y}^{2} = 2x {y}^{3} [/tex]

c)

[tex]6b \times ( - a {)}^{2} \times ( - \frac{1}{2} ) {a}^{3} b = 6b {a}^{2} \times ( - \frac{1}{2} ) {a}^{3} b = - 3b {a}^{2} \times {a}^{3} b = - 3 {a}^{5} {b}^{2} [/tex]