p^2q-(q^2p-rq-(qr-(p^2q+q^2r)+(q^3-p^3)))+p^3-q^3-r^3 if p=1q=-1r=2 find the value
Answers
Answered by
1
2p3q5r=p2q3r52p3q5r=p2q3r5
only possible when p=2,q=3,r=5p=2,q=3,r=5
so p+q+r=2+3+5=10p+q+r=2+3+5=10
For those who like detailed answer :
2p3q5rp2q3r5=12p3q5rp2q3r5=1
2pp2.3qq3.5rr5=12pp2.3qq3.5rr5=1
Each fraction needs to be equal to 1 because considering any other value will give you only a relation among fractions.
2pp2=12pp2=1
2p=p22p=p2
Taking log both sides,
2logp=plog22logp=plog2
logpp=log22logpp=log22
p1p=212p1p=212
By comparing p=2p=2, Similarly, q=3q=3 and r=5r=5
p+q+r=10
pls mark as brainlist
only possible when p=2,q=3,r=5p=2,q=3,r=5
so p+q+r=2+3+5=10p+q+r=2+3+5=10
For those who like detailed answer :
2p3q5rp2q3r5=12p3q5rp2q3r5=1
2pp2.3qq3.5rr5=12pp2.3qq3.5rr5=1
Each fraction needs to be equal to 1 because considering any other value will give you only a relation among fractions.
2pp2=12pp2=1
2p=p22p=p2
Taking log both sides,
2logp=plog22logp=plog2
logpp=log22logpp=log22
p1p=212p1p=212
By comparing p=2p=2, Similarly, q=3q=3 and r=5r=5
p+q+r=10
pls mark as brainlist
Similar questions