Question

Someone please help....​

Answer 1

Answer:

x=9

Step by step explanation:

$\frac{x+3}{x-3} +\frac{x-3}{x+3} =2\frac{1}{2} \\\\L.C.M\\\\=\frac{(x+3)^2+(x-3)^2}{x^2-9}=2\frac{1}{2} \\\\=\frac{x^2+9+6x+x^2+9-6x}{x^2-9}=2\frac{1}{2}\\\\=\frac{2x^2+18}{x^2-9}=2\frac{1}{2}\\\\=\frac{2(x^2+9)}{x^2-9}=\frac{5}{2}\\\\~Cross~multiplying:\\\\=4x^2+36 = 5x^2-45 \\\\=x^2=81\\\\x=9$

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Answer 2

Answer:

this is your answer sis