Question

plz slv it friends. ​

Answer 1

Step-by-step explanation:

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$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}$

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$\implies \frac{1}{x + 1} + \frac{2}{ x + 2 } = \frac{4}{x + 4} \\ \implies \frac{x + 2 + 2x + 2}{ {x}^{2} + 3x + 2} = \frac{4}{x + 4} \\ \implies \frac{3x + 4}{(x + 1)(x + 2)} - \frac{4}{x + 4} = 0 \\ \implies \frac{(3x + 4)(x + 4) - 4( {x}^{2} + 3x + 2) }{(x + 1)(x + 2)(x + 4)} = 0 \\ \implies \frac{3{x}^{2} +16x +16 - 4{x}^{2} -12x -8}{(x + 1)(x + 2)(x + 4)} = 0 \\ \implies \frac{ -{x}^{2} +4x +8}{(x + 1)(x + 2)(x + 4)} = 0 \\ \implies {x}^{2} - 4x -8 =0 \\ \implies x = \frac{4 + \: or \: - \sqrt{16 +32} }{2} \\ \implies x = \frac{4 + \: or \: - \sqrt {48} }{2} \\ \implies x = \frac{4 + \: or \: - 4\sqrt{3} }{2} \\ \implies \boxed { x = 2 + 2\sqrt{3} \; \; or \; \; x = 2 - 2\sqrt{3} }$