Question

p=x÷x+y, q=y÷x+y, then find 1÷p-q- 2q÷p2-q2

Answer 1

Answer:

1

Step-by-step explanation:

p = x/(x+y)

q = y/(x+y)

To find

1/(p-q) - 2q/(p²-q²)

=1/(p-q) - 2q/{(p+q)(p-q)}

=(1/(p+q)(p-q))(p + q - 2q)

= (1/((p-q)(p+q))(p - q)

= 1/p+q

p + q = x/(x+y) + y/(x+y)

=(x+y)/(x+y)

= 1

1/(p-q) - 2q/(p²-q²) = 1/1

=> 1/(p-q) - 2q/(p²-q²) = 1

Answer 2

Answer:

The solution of the expression 1÷p-q- 2q÷p²-q² is 1.

Step-by-step explanation:

Given :

• p = x/ (x+y)
• q = y/ (x+y)

Here, p-q = x/(x+y) - y/(x+y)

                = (x-y)/(x+y)

p²-q² = (p+q)(p-q)

         = [(x+y)/(x+y)][(x-y)/(x+y)]

p+q = x/(x+y) + y/(x+y)

                = (x+y)/(x+y)

                = 1

Therefore, simplifying the given expression

(1/p-q) - (2q/p²-q²) = (p+q-2q)/p²-q²

                               = (p-q)/(p+q)(p-q)

                                = 1/p+q

                                 = 1

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