Question

1/(x-y) + 1 /(x + y) + 2x/(x^2+y^2) + 4x^3/(x^4+y^4)​

Answer 1

Answer:

8x^7 / x^8 - y^8

Step-by-step explanation:

1. 1/(x-y) + 1 /(x + y)

= x + y + x - y / (x-y)(x+y)

= 2x / (x^2 - y^2)

2.  2x / (x^2 - y^2) + 2x / (x^2 + y^2)

= 2x^3 + 2xy^2 + 2x^3 - 2xy^2 / (x^4 - y^4)

= 4x^3 / (x^4 - y^4)

3.   4x^3 / (x^4 - y^4) + 4x^3/(x^4+y^4)​

= 4x^7 + 4x^3y^4 + 4x^7 - 4x^3y^4

= 8x^7 / (x^8 - y^8)

Answer 2

$\large\mathscr \color{black}\colorbox{red}{\colorbox{orange}{\colorbox{yellow}{\colorbox{green}{\colorbox{blue}{\colorbox{indigo}{\colorbox{violet}{\underline {ANSWER}}}}}}}}$

$\large\tt\color{lime}{ \frac{1}{x - y} + \frac{1}{x + y} } + \frac{2x}{ {x}^{2} + {y}^{2} } + \frac{ {4x}^{3} }{ {x}^{4} + {y}^{4} } \\ \\\longmapsto\large\tt\color{lime}{ \frac{(x + y) + (x - y)}{ {x}^{2}- {y}^{2} } + \frac{2x}{ { {x}^{2} } + {y}^{2} } + \frac{ {4x}^{3} }{ {x}^{4} + {y}^{4} } } \\ \\ \longmapsto\large\tt\color{lime}{\frac{2x}{ {x}^{2} - {y}^{2}} + \frac{2x}{ {x}^{2} + {y}^{2} } + \frac{4 {x}^{3} }{ {x}^{4} + {y}^{4} } }$

$\longmapsto\large\tt\color{lime}{ \frac{2x( { {x}^{2} + y}^{2} + {x}^{2} - {y}^{2}) }{ ({x}^{2} - {y}^{2})( {x}^{2} + {y}^{2} ) } } + \frac{4 {x}^{3} }{ {x}^{4} + {y}^{4} } \\ \\\longmapsto \large\tt\color{lime}{\frac{4 {x}^{3} }{ {x}^{4} - {y}^{4} } + \frac{4 {x}^{ 3} }{ {x}^{4} + {y}^{4} } }\\\\\longmapsto\large\tt\color{lime} \frac{4 {x}^{3} ( {x}^{4} + {y}^{4} + {x}^{4} - {y}^{4} )}{( {x}^{4} + {y}^{4})( {x}^{4} - {y}^{4})}\\\\\longmapsto\large\tt\color{cyan}{\underline{\underline{\frac{{8x}^{7}}{{x}^{8}-{y}^{8}}}}}$