Question

Pls help me to solve 19th sum guys pls help me

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

$\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{5}{x + 4} \\ \\ \frac{x + 2 + 2(x + 1)}{(x + 1)(x + 2)} = \frac{5}{x + 4} \\ \\ \frac{x + 2 + 2x + 2}{ {x}^{2} + 2x + + x + 2 } = \frac{5}{x + 4} \\ \\ \frac{3x + 4}{ {x}^{2} + 3x + 2 } = \frac{5}{x + 4} \\ \\ {(3x + 4)(x + 4)} = 5( {x}^{2} + 3x + 2) \\ \\ {3x}^{2} + 12x + 4x + 16 = {5x}^{2} + 15x + 10 \\ \\ {3x}^{2} + 16x + 16 = {5x}^{2} + 15x + 10 \\ \\ {3x}^{2} - {5x}^{2} + 16x - 15x + 16 - 10 = 0 \\ \\ - 2 {x}^{2} + x + 6 = 0 \\ \\ 2 {x}^{2} - x - 6 = 0 \\ \\ 2 {x}^{2} - 4x + 3x - 6 =0 \\ \\ 2x(x - 2) + 3(x - 2) = 0 \\ \\ (2x + 3)(x - 2) = 0 \\ \\ x = 2..........or \: ............x = \frac{ - 3}{2}$

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