Question

Please solve it I have exam

Answer 1

Answer:

x = 1

Step-by-step explanation:

⇒ $\frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} = \frac{1}{2a+b+x}$

⇒ $\frac{2bx+4ax+2ab}{4abx} = \frac{1}{2a+b+x}$

⇒ (2bx + 4ax + 2ab)(2a + b + x) = 4abx

⇒ 4abx + 8a²x + 4a²b + 2b²x + 4abx + 4ax² + 2bx² + 4ax² + 2abx = 4abx

⇒ 8ax(a+x) + 4a²b + 2bx(b+x) + 6abx = 0

⇒ [8a + 2b]x² + [8a² + 6ab + 2b²]x + 4a²b = 0

⇒ x = {-(8a² + 6ab + 2b²) ± √(8a² + 6ab + 2b²)² - 16a²b(8a+2b)} / {2(8a+2b)}

⇒ x = 1