The simplest form of the expression $$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}}$$ + $$\frac{{{p^2} - 1}}{{{p^2} + 3p}}$$ + $$\frac{{{p^2}}}{{p + 1}}$$ = ?
A. 2p3
B. $$\frac{1}{{2{p^2}}}$$
C. p + 3
D. $$\frac{1}{{p + 3}}$$
Answer: Option B
Solution(By Examveda Team)
$$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}} + \frac{{{p^2} - 1}}{{{p^2} + 3p}} + \frac{{{p^2}}}{{p + 1}}$$In such type of question assume values of p
$$\eqalign{ & \therefore {\text{Let }}p = 1 \cr & \therefore \frac{{1 - 1}}{{2 + 1}} + \frac{{1 - 1}}{{1 + 3}} + \frac{1}{{1 + 1}} \cr & = 0 + 0 + \frac{1}{2} \cr & = \frac{1}{2} \cr & {\text{Now check option 'B'}} \cr & \frac{1}{{2{p^2}}} = \frac{1}{2} \cr & {\text{Hence the answer is option 'B'}} \cr} $$
Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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