Question

please please please please please please answer fast​

Answer 1

Answer:

( i ) zy-xy ( ii ) 5a²b² - 12ab +8. I've been answered your questions so please mark me as a brainlist and please follow me on and please thank me if this answer is helpful to you and

Step-by-step explanation:

Given : Add

(i ) yx - yz , zy - zx , xz-xy

sol add

(yx - yz) + (zy - zx ) + (xz - xy)

removing brackets we get

yz -yz +zy -zx + xz -xy

taking common terms aside and cancel it

yz -yz cancel next 1 zy left next +xz- xz cancel we get alast after cancellation is

= zy-xy

(ii)

$2 {a}^{2} {b}^{2} - 3ab + 5 \: and \: 3 - 9ab + 3 {a}^{2} {b}^{2}$

(2a² b² - 3ab +5 ) + ( 3-9ab +3a²b²)

= 2a²b² -3ab +5 + 3-9ab +3 a²b²

= taking common terms aside and cancel it

= 2a²b² +3a²b² -3ab - 9ab +5 +3

= now add or subtract the common terms

= 5a²b² - 12ab +8