Answer c plzz wanted urgently
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"l" is common in the first set,& "y" is common in second set.so,
l(x+m)-y(x+m)
(x+m) is common in this case so taking it separately gives
(l-y)(x+m)
l(x+m)-y(x+m)
(x+m) is common in this case so taking it separately gives
(l-y)(x+m)
Answered by
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A) mx - my + lx -ly
= m (x-y) +l (x-y)
=(x-y)(m+l)
B) lxy + mxy -mz-lz
= xy (l+m) -z (l+m)
= (l+m)(xy-z)
C) lx + lm - xy - my
= l (x+m) -y (x+m)
= (x+m)(l-y)
D) lx + l^2y -mx -lmy
= l (x+ly) -m (x+ly)
= (x+ly)(l-m)
HOPE THIS HELPS YOU! ! ! !
= m (x-y) +l (x-y)
=(x-y)(m+l)
B) lxy + mxy -mz-lz
= xy (l+m) -z (l+m)
= (l+m)(xy-z)
C) lx + lm - xy - my
= l (x+m) -y (x+m)
= (x+m)(l-y)
D) lx + l^2y -mx -lmy
= l (x+ly) -m (x+ly)
= (x+ly)(l-m)
HOPE THIS HELPS YOU! ! ! !
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