Question

(v) 5pq(p2 - q2) / 2p(p+q)
question is 5 pq bracket me p ka square +q ka square divided by 2 p ( p+ q)​

Answer 1

Given Question :- Simplify

$\rm \: \dfrac{5pq( {p}^{2} - {q}^{2})}{2p(p + q)}$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \dfrac{5pq( {p}^{2} - {q}^{2})}{2p(p + q)}$

Step :- 1 Cancel out p

$\rm \: = \: \dfrac{5 \cancel{p}q( {p}^{2} - {q}^{2})}{2\cancel{p}(p + q)}$

So, we get now

$\rm \: = \: \dfrac{5q( {p}^{2} - {q}^{2})}{2(p + q)}$

Step :- 2

We know,

$\boxed{ \rm{ \: {x}^{2} - {y}^{2} = (x + y)(x - y) \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{5q(p + q)(p - q)}{2(p + q)}$

$\rm \: = \: \dfrac{5q \: \cancel{(p + q)} \: (p - q)}{2 \: \cancel{(p + q)} \: }$

$\rm \: = \: \dfrac{5}{2} \: q \: (p - q)$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{5pq( {p}^{2} - {q}^{2})}{2p(p + q)} = \frac{5}{2}q(p - q) \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Question :-

$\leadsto \sf \dfrac{5pq(p^2 - q^2)}{2p(p + q)}$

Solution :-

Given :

$\dashrightarrow \bf \dfrac{5pq(p^2 - q^2)}{2p(p + q)}$

First, we have to break (p² - q²) by using algebraic identities :

As we know that :

$\bigstar \: \: \sf\boxed{\bold{\pink{a^2 - b^2 =\: (a + b)(a - b)}}}\: \: \: \bigstar$

So, by taking help with this algebraic identities we get,

$\longrightarrow \sf \dfrac{5pq(p^2 - q^2)}{2p(p + q)}$

$\dashrightarrow \sf \dfrac{5pq(p + q)(p - q)}{2p(p + q)}$

Second, by cancel (p + q) :

$\dashrightarrow \sf \dfrac{5pq\cancel{(p + q)}(p - q)}{2p\cancel{(p + q)}}$

$\dashrightarrow \sf \dfrac{5pq(p - q)}{2p}$

Third, again by cancel p :

$\dashrightarrow \sf \dfrac{5\cancel{p}q(p - q)}{2\cancel{p}}$

$\dashrightarrow \sf \dfrac{5q(p - q)}{2}$

$\dashrightarrow \sf \dfrac{5 \times q \times (p - q)}{2}$

$\dashrightarrow \sf \dfrac{5}{2} \times q \times (p - q)$

$\dashrightarrow \sf\bold{\red{\dfrac{5}{2}q(p - q)}}$

$\sf\bold{\underline{\purple{\therefore\: The\: value\: of\: \dfrac{5pq(p^2 - q^2)}{2p(p + q)}\: is\: \dfrac{5}{2}q(p - q)\: .}}}$