Question

(2/3a²b²c² + 4/3 ab²c²) ÷ 1/2 abc, Divide the given polynomial by the given monomia

Answer 1

Solution :

( 2/3 a²b²c² + 4/3 ab²c² )÷ 1/2abc

= [(2/3a²b²c²)/(1/2 abc )+[ 4/3ab²c²)/(1/2abc)]

After cancellation , we get

= (4/3)abc + (8/3)bc

•••

Answer 2

$= > \dfrac{( \frac{2}{3} {a}^{2} {b}^{2} {c }^{2} + \frac{4}{3} a {b}^{2} {c}^{2} ) }{ \frac{1}{2} abc} \\ \\ \\ = > \dfrac{ \frac{2 {a}^{2} {b}^{2} {c}^{2} }{3} + \frac{4a {b}^{2} {c}^{2} }{3} }{ \frac{abc}{2} } \\ \\ \\ = > \dfrac{ \frac{2{a}^{2} {b}^{2} {c}^{2} + 4 {ab}^{2} {c}^{2} }{3} }{ \frac{abc}{2} } \\ \\ \\ = > \frac{2{a}^{2} {b}^{2} {c}^{2} + 4 {ab}^{2} {c}^{2} }{3} \times \frac{2}{abc} \\ \\ \\ = > \frac{abc(2abc + 4bc)}{3} \times \frac{2}{abc} \\ \\ \\ = > \frac{2abc + 4bc}{3} \times 2$



= > ( 4abc + 8bc ) / 3