Question

bài 1 ; phép nhân

a) 3a² ( a² - 4a + 6 )

b) (2b² - 1/3b + 4 ) ( -3/6b³ )

Answer 1

a: \(3a^2\left(a^2-4a+6\right)=3a^2\cdot a^2-3a^2\cdot4a+3a^2\cdot6\)

\(=3a^4-12a^3+18a^2\)

b: \(\left(2b^2-\dfrac{1}{3}b+4\right)\cdot\left(-\dfrac{3}{6}b^3\right)\)

\(=-\dfrac{1}{2}b^3\left(2b^2-\dfrac{1}{3}b+4\right)=-\dfrac{1}{2}b^3\cdot2b^2+\dfrac{1}{2}b^3\cdot\dfrac{1}{3}b-\dfrac{1}{2}b^3\cdot4\)

\(=-b^5+\dfrac{1}{6}b^4-2b^3\)

Answer 2

________

\(a.3a^2\left(a^2-4a+6\right)\\ =3a^4-12a^3+18a^2\\ b.\left(2b^2-\dfrac{1}{3}b+4\right)\left(-\dfrac{3}{6}b^3\right)\\ =\left(2b^2-\dfrac{1}{3}b+4\right)\left(-\dfrac{1}{2}b^3\right)\\ =-b^5+\dfrac{1}{6}b^4-2b^3\)