Question

if x = 2mn/m+n show that x+m/x-m + x+n/x-n = 2​

Answer 1

Answer:

sorry I can't answer it

Answer 2

Given :

• $\\ \textsf {if x =} {\tt \: \dfrac{2 \: m \: n}{m + n}} \: \: \textsf{show that } \: \: \tt{ \dfrac{x + m}{x - m} + \dfrac{x + n}{x - n} = 2} \:$

Solution :

$\\ \tt x = \frac{2mn}{m + n} \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \tt\frac{x }{m} = \frac{2n}{m + n}$

By applying the formula of Componendo and Dividendo in the equation we get :

$\\ \tt \: \: \: \dashrightarrow \: \frac{x + m}{x - m} = \frac{2n + m + n}{2n - m - n} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \tt \: \dashrightarrow \: \: \: \frac{x + m}{x - m} = \frac{3n + m }{n - m } \: \: ——— \:( 1)$

Again,

$\\ \tt x = \frac{2mn}{m + n} \: \: \: \: \: \: \: \: \: \\ \\ \\ : \implies \tt\frac{x }{n} = \frac{2m}{m + n}$

By applying the formula of Componendo and Dividendo in the equation we get :

$\\ \tt \: \: \: \dashrightarrow \: \frac{x + n}{x - n} = \frac{2m + m + n}{2m - m - n} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \tt \: \dashrightarrow \: \: \: \frac{x + n}{x - n} = \frac{3m+ n }{m - n } \: \: ——— \:( 2)$

Now, adding equation (1) and (2) :

$\\ \tt \: \frac{x + m}{x - m} + \frac{x + n}{ x- n} = \frac{3n + m}{n - m} + \frac{3m + n}{m - n} \\ \\ \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{3n + m}{n - m} - \frac{3m + n}{n - m} \\ \\ \\ \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:= \frac{3n + m - 3m - n}{n - m} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \:\tt \: = \: \: \frac{2n - 2m}{n - m} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \tt \: = \: \frac{2 \cancel {(n - m)}}{ \cancel{(n - m)}} \\ \\ \\ \tt \: = \: { \mathfrak{ \pmb{2}}}$