Question

please help me to solve this ​

Answer 1

Answer:

$\large{\bold{\frac{4(x-1)}{3(x-4)(x-2)}}}$

Step-by-step explanation:

$Given\\\\\frac{x+2}{x^2+5x+4}+\frac{x-4}{3(x^2-3x+2)}\\\\Consider\\x^2-5x+4\\\\=x^2-4x-x+4\\\\=x(x-4)-(x-4)\\\\=(x-1)(x-4)\\\\Consider\\x^2-3x+2\\\\=x^2-2x-x+2\\\\=x(x-2)-(x-2)\\\\=(x-1)(x-2)\\\\\frac{x+2}{x^2+5x+4}+\frac{x-4}{3(x^2-3x+2)}\\\\=\frac{x+2}{(x-1)(x-4)}+\frac{x-4}{3(x-1)(x-2)}\\\\=\frac{1}{(x-1)}(\frac{x+2}{x-4}+\frac{x-4}{x-2})\\\\=\frac{1}{(x-1)}(\frac{3(x+2)(x-2)+(x-4)(x-4)}{3(x-4)(x-2)})\\\\=\frac{1}{(x-1)}(\frac{3(x^2-4)+(x^2-8x+16)}{3(x-4)(x-2)})$

$=\frac{1}{(x-1)}(\frac{4x^2-8x+4}{3(x-4)(x-2)})\\\\=\frac{1}{(x-1)}(\frac{4(x^2-2x+1)}{3(x-4)(x-2)})\\\\=\frac{1}{(x-1)}(\frac{4(x-1)^2}{3(x-4)(x-2)})\\\\=\frac{4(x-1)}{3(x-4)(x-2)}$

HOPE IT HELPS