simplify: x^2+6x+5/x2-25?
回答 (8)
2008-12-27 2:45 pm
✔ 最佳答案
= (x² + 6x + 5)/(x² - 25)= ([x + 5][x + 1])/([x + 5][x - 5])
= (x + 1)/(x - 5)
Answer: (x + 1)/(x - 5)
Checking:
= ([x + 1]/[x - 5])(x² - 25)
= ([x + 1]/[x - 5])([x + 5][x - 5])
= (x + 1)(x + 5)
= x² + 5x + x + 5
= x² + 6x + 5
2008-12-27 7:14 pm
(x^2 + 6x + 5)/(x^2 - 25)
= (x^2 + 5x + x + 5)/(x^2 + 5x - 5x - 25)
= (x + 5)(x + 1)/(x + 5)(x - 5)
= (x + 1)/(x - 5)
= (x^2 + 5x + x + 5)/(x^2 + 5x - 5x - 25)
= (x + 5)(x + 1)/(x + 5)(x - 5)
= (x + 1)/(x - 5)
2008-12-27 10:41 pm
x^2+6x+5/ x^2-25
X+1 x+5 / x+5 x-5
x+1/x-5
X+1 x+5 / x+5 x-5
x+1/x-5
2008-12-27 7:18 pm
(x+1)/(x-5)
2008-12-27 9:58 pm
x^2+6x+5/x2-25
now factorize x^2+6x+5
x^2 + 5x + x + 5 =0
x(x+5) + 1(x+5)
therefore
(x+5)(x+1)
and for
x^2-25
x^2 - 5^2
using a^2 - b^2 = (a+b)(a-b)
= (x-5)(x+5)
now
(x+5)(x+1) / (x-5)(x+5)
= (x+1) / (x-5)
now factorize x^2+6x+5
x^2 + 5x + x + 5 =0
x(x+5) + 1(x+5)
therefore
(x+5)(x+1)
and for
x^2-25
x^2 - 5^2
using a^2 - b^2 = (a+b)(a-b)
= (x-5)(x+5)
now
(x+5)(x+1) / (x-5)(x+5)
= (x+1) / (x-5)
2008-12-27 9:44 pm
scratches chin......ok<wink>
x^2+6x+5 /(x^2-25) =[ (x^2-25)+(6x+30)] / (x^2-25)
= 1 + 6x+30/(x^2-25) = 1 + 6(x+5)/(x+5)(x-5) = 1+ 6/(x-5)
= x+1 / x-5
(ofcourse there are other approaches..........)
x^2+6x+5 /(x^2-25) =[ (x^2-25)+(6x+30)] / (x^2-25)
= 1 + 6x+30/(x^2-25) = 1 + 6(x+5)/(x+5)(x-5) = 1+ 6/(x-5)
= x+1 / x-5
(ofcourse there are other approaches..........)
2008-12-27 7:37 pm
x^2+6x+5/x^2-25
=(x+5)(x+1)
----------------- cancel x+5
(x+5)(x-5)
= x+1
--------- answer @(^_^)@
x-5
=(x+5)(x+1)
----------------- cancel x+5
(x+5)(x-5)
= x+1
--------- answer @(^_^)@
x-5
2008-12-27 7:31 pm
Factor numerator and denominator and cancel equal factors.
(x² + 6x +5) / (x² - 25)
= (x + 1)(x + 5) / (x + 5)(x - 5)
= (x + 1) / (x - 5)
(x² + 6x +5) / (x² - 25)
= (x + 1)(x + 5) / (x + 5)(x - 5)
= (x + 1) / (x - 5)
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