Question

can anyone solve this pleaseeeee

Answer 1

Assume (p/q)=x

So. (p/q)+(q/p)=2

i.e., x+(1/x)=2

x^2 + 1 = 2x

x^2 -2x +1 =0

x^2-x-x+1=0

x(x-1)-1(x-1)=0

(x-1)(x-1)=0

(x-1)=0 or (x-1)=0

x=1 and x=1

･ﾟ･ (p/q)=(q/p)=1

Consider ,

(p/q)^23+(q/p)^7

=(1)^27+(1)^7

=1+1

=2

Thank you

Answer 2

$\bigg( \dfrac{p}{q}\bigg)^{23} + \bigg( \dfrac{q}{p} \bigg)^{7}$

Define x:

$\text {Let x = } \dfrac{p}{q}$

$\text {Therefore: } \dfrac{q}{p} = \dfrac{1}{x}$

Write in term of x:

$\dfrac{p}{q} + \dfrac{q}{p} = 2$

$x + \dfrac{1}{x} = 2$

$\dfrac{x^2 + 1}{x} = 2$

$x^2 + 1 = 2x$

$x^2 -2x + 1 = 0$

$(x - 1)^2 = 0$

$x - 1 = 0$

$x = 1$

Find p/q:

$\dfrac{p}{q} = x$

$\dfrac{p}{q} = 1$

Find q/p:

$\dfrac{q}{p} = \dfrac{1}{x}$

$\dfrac{q}{p} = 1$

Find (p/q)²³ + (q/p)⁷:

$\bigg( \dfrac{p}{q} \bigg)^{23} + \bigg( \dfrac{q}{p} \bigg)^{7} = (1)^{23} + (1)^7 = 2$

Answer: 2