Respuesta :

beinggreat78

Answer:

[tex]\mathrm{\frac{(a^7+a^6+a^5+a^4+a^3+a^2+a+1)}{((a+1)*(a^2+1)*(a^4+1)*(a-1))} }[/tex]

Step-by-step explanation:

[tex](A+B) * (A-B) =\\ A^2 - AB + BA - B^2 =\\ A^2 - AB + AB - B^2 =\\ A^2 - B^2\\-AB + AB\\= 0\\1 = \sqrt{1} \\a^8 = \sqrt{a^4} \\= a^4 + 1 * a^4 - 1\\a^4 - 1\\a^4 = \sqrt{a^2}\\= a^2 + 1 * a^2 - 1\\a^2 + 1\\a^2 = \sqrt{a^1}\\= a + 1 * a - 1\\=\frac{1}{a+1} + \frac{2}{a^2 + 1} + \frac{4}{a^4 + 1} + \frac{8}{(a^4 + 1) * (a^2 + 1) * (a + 1) * (a - 1)}\\\\\mathrm{Remove\:parentheses\:if\:any}[/tex]

[tex]= \frac{1}{a+1} + \frac{2}{a^2 + 1} + \frac{4}{a^4 + 1} + \frac{8}{a^4 + 1 * a^2 + 1 * a + 1 * a - 1}\\\\\mathrm{LCMs}\\a + 1\\a^2 + 1\\a + 1 * a^2 + 1\\\mathrm{Left: a^2 + 1}\\\mathrm{Right: a + 1}\\\\\mathrm{Left} = \frac{a^2 + 1}{a + 1 * a^2 + 1}\\\mathrm{Right} = \frac{2 * a + 1}{a + 1 * a^2 + 1}\\Combine\\= \frac{a^2 + 1 + 2 * (a+1)}{a + 1 * a^2 + 1}\\= \frac{a^2 + 2a + 3}{a + 1 * a^2 + 1}\\= \frac{a^2 + 2a + 3}{a + 1 * a^2 + 1} + \frac{4}{a^4 + 1}\frac{4}{a^4 + 1} +[/tex]

[tex]\frac{8}{a^4 + 1 * a^2 + 1 * a + 1 * a - 1}\\\\a^2 + 2a + 3\\(a+1) * (a^2+1) * (a^4+1)\\\mathrm{LCMs}\\=a^4+1\\=(a+1)*(a^2+1)\\= \frac{(a^2+2a+3) * (a^4+1)}{(a+1) * (a^2+1) * (a^4+1)}\\= \frac{4 * (a+1) * (a^2+1)}{(a+1) * (a^2+1) * (a^4+1)}\\= \frac{(a^2+2a+3) * (a^4+1) + 4 * (a+1) • (a^2+1)}{(a+1) * (a^2+1) * (a^4+1)}\\ = \frac{a^6 + 2a^5 + 3a^4 + 4a^3 + 5a^2 + 6a + 7 }{(a + 1) * (a^2 + 1) * (a^4 + 1)}\\ = \frac{a^6 + 2a^5 + 3a^4 + 4a^3 + 5a^2 + 6a + 7 }{(a + 1) * (a^2 + 1) * (a^4 + 1)}[/tex]

[tex]+ \frac{8}{a^4 + 1 * a^2 + 1 * a + 1 * a - 1}\\\\\mathrm{LCMs}\\ (a+1) * (a^2+1) * (a^4+1) * (a-1)\\= a-1\\= 1\\= \frac{(a^6+2a^5+3a^4+4a^3+5a^2+6a+7) * (a-1)}{(a+1) * (a^2+1) * (a^4+1) * (a-1)} + \frac{8}{a^4 + 1 * a^2 + 1 * a + 1 * a - 1}\\= \frac{a^7 + a^6 + a^5 + a^4 + a^3 + a^2 + a + 1}{(a + 1) * (a^2 + 1) * (a^4 + 1) * (a - 1)}[/tex]

I really hope this helps, this took me a very long time to figure out. If you have concerns, please feel free to address them to me.

mhanifa

Answer:

  • [tex]\cfrac{ 1}{a-1}[/tex]

========================

Use identity

  • x² - y² = (x + y)(x - y)

Start simplification form the end backwards

[tex]\cfrac{4}{a^4+1}+ \cfrac{8}{a^8-1}=[/tex]

[tex]\cfrac{4(a^4-1)}{(a^4+1)(a^4-1)}+\cfrac{8}{(a^4+1)(a^4-1)}=[/tex]

[tex]\cfrac{4a^4-4+8}{(a^4+1)(a^4-1)}=[/tex]

[tex]\cfrac{4a^4+4}{(a^4+1)(a^4-1)}=[/tex]

[tex]\cfrac{4(a^4+1)}{(a^4+1)(a^4-1)}=[/tex]

[tex]\cfrac{4}{a^4-1}[/tex]

Add the second term

[tex]\cfrac{2}{a^2+1} +\cfrac{4}{a^4-1}=[/tex]

[tex]\cfrac{2(a^2-1)}{(a^2+1)(a^2-1)} +\cfrac{4}{(a^2+1)(a^2-1)}=[/tex]

[tex]\cfrac{2a^2-2+4}{(a^2+1)(a^2-1)}=[/tex]

[tex]\cfrac{2a^2+2}{(a^2+1)(a^2-1)}=[/tex]

[tex]\cfrac{2(a^2+1)}{(a^2+1)(a^2-1)}=[/tex]

[tex]\cfrac{2}{a^2-1}[/tex]

Add the first term

[tex]\cfrac{1}{a+1}+\cfrac{2}{a^2-1}=[/tex]

[tex]\cfrac{a - 1}{(a+1)(a-1)}+\cfrac{2}{(a+1)(a-1)}=[/tex]

[tex]\cfrac{a - 1+2}{(a+1)(a-1)}=[/tex]

[tex]\cfrac{a + 1}{(a+1)(a-1)}=[/tex]

[tex]\cfrac{ 1}{a-1}[/tex]

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