KeerthanaPosted on If p=3+1pp=3+\frac{1}{\mathrm{p}}p=3+p1, then the value of 5pp2+p−1\frac{5 \mathrm{p}}{\mathrm{p}^{2}+\mathrm{p}-1}p2+p−15p is : a1141 \frac{1}{4}141 b1 c3123 \frac{1}{2}321 d2132 \frac{1}{3}231 Answer : Option AExplanation : p−1p=3\mathrm{p}-\frac{1}{\mathrm{p}}=3p−p1=3 (Given) Expression =5pp2+p−1=\frac{5 p}{p^{2}+p-1}=p2+p−15p =5pp(p+1−1p)=5(p−1p)+1=53+1=54=114=\frac{5 p}{p\left(p+1-\frac{1}{p}\right)}=\frac{5}{\left(p-\frac{1}{p}\right)+1}=\frac{5}{3+1}=\frac{5}{4}=1 \frac{1}{4}=p(p+1−p1)5p=(p−p1)+15=3+15=45=141 Rate This:NaN / 5 - 1 votesAdd comment
