5n+3−6× 5n+19 × 5n−5n × 22 = 19
True
Given: 5n+3−6× 5n+19 × 5n−5n × 22
= 5n+1+2−6× 5n+19 × 5n−5n × 22
= (5n+1×52)−6× 5n+19 × 5n−5n × 22 (∵am×an = a(m+n))
Taking the common factor out we get
5n+1(52 − 6)5n(9 − 22)
⇒ 5n+1(19)5n(5)
⇒ 5n+1(19)5n+1
⇒ 19