Question

pls solve this for me pls​

Answer 1

HELLO MATE......

Given........

$t1 = \frac{1}{x + 2} \\ \\ t2 = \frac{1}{x + 3} \\ \\ t3 = \frac{1}{x + 5} \\ \\ we \: know \: = > \\ \\ t2 - t1 = t3 - t2 \\ \\ sub \: that \: values \: = > \\ \\ \frac{1}{x + 3} - \frac{1}{x + 2} = \frac{1}{x + 5} - \frac{1}{x + 3} \\ \\ \frac{(x + 2) - (x + 3)}{(x + 3)(x + 2)} = \frac{(x + 3) - (x + 5)}{(x + 3)(x + 5)} \\ \\ \frac{x + 2 - x - 3}{x + 2} = \frac{x + 3 - x - 5}{x + 5} \\ \\ \frac{ - 1}{x + 2} = \frac{ - 2}{x + 5} \\ \\ - x - 5 = - 2x - 4 \\ \\ 2x - x = - 4 + 5 \\ \\ x = 1$

Hope it helps u

Answer 2

Step-by-step explanation:

good night...

Mark me brain list and have a great day ahead ....