Question

x+2/x-2 - 6/x+2 = x²/x²+4

Answer 1

Sửa đề: \(\dfrac{x+2}{x-2}-\dfrac{6}{x+2}=\dfrac{x^2}{x^2-4}\)

\(\Leftrightarrow x^2+4x+4-6x+12=x^2\)

=>-2x+16=0

hay x=8(nhận)

Answer 2

Ta có:\(\dfrac{x+2}{x-2}\)- \(\dfrac{6}{x+2}\)= \(\dfrac{x^2}{x^2-4}\)

ĐKXĐ: x ≠ \(\pm\)2

⇔ \(\dfrac{\left(x+2\right)\left(x+2\right)-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)= \(\dfrac{x^2}{\left(x+2\right)\left(x-2\right)}\)

⇔ (x + 2)(x + 2) - 6(x - 2) = \(x^2\)

⇔ x² + 2x + 2x + 4 - 6x + 12 = x²

⇔ 4x + 4 - 6x + 12 = 0

⇔ 16 - 2x = 0

⇔ x = 8 (thỏa mãn)

Vậy S = {8}