神马作文网一道2元2次的提怎么约分
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一道2元2次的提怎么约分
6aq=a+a(q^2)+7
[a(1-q^3)]/(1-q)=7
就这2式子.如何约分.我就这点分了.
6aq=a+a(q^2)+7
[a(1-q^3)]/(1-q)=7
就这2式子.如何约分.我就这点分了.
6aq=a+a(q^2)+7
6aq - a - aq^2 = 7
a(6q-1-q^2) =7
[a(1-q^3)]/(1-q)=7
(a(1-q^3))/(1-q)=7
a(1-q)(1+q+q^2)/（1-q） =7
a(1+q+q^2) = 7
a(6q-1-q^2) = a(1+q+q^2)
a(6q-1-q^2-1-q-q^2) =0
a(5q-2-2q^2) = 0
a = 0
当a = 0
q无解~~设去
5q-2-2q^2 = 0
q= 1/2
q = 2
当q= 1/2
6aq=a+a(q^2)+7
3a = a + 0.25a +7
1.75a = 7
a = 4
当q = 2
6aq=a+a(q^2)+7
12a = a + 4a+7
7a=7
a= 1
6aq - a - aq^2 = 7
a(6q-1-q^2) =7
[a(1-q^3)]/(1-q)=7
(a(1-q^3))/(1-q)=7
a(1-q)(1+q+q^2)/（1-q） =7
a(1+q+q^2) = 7
a(6q-1-q^2) = a(1+q+q^2)
a(6q-1-q^2-1-q-q^2) =0
a(5q-2-2q^2) = 0
a = 0
当a = 0
q无解~~设去
5q-2-2q^2 = 0
q= 1/2
q = 2
当q= 1/2
6aq=a+a(q^2)+7
3a = a + 0.25a +7
1.75a = 7
a = 4
当q = 2
6aq=a+a(q^2)+7
12a = a + 4a+7
7a=7
a= 1