how do you solve (2m-n)^4??
回答 (6)
2008-04-16 3:56 pm
✔ 最佳答案
you either do it by (2m-n)*(2m-n)*(2m-n)*(2m-n) (multiplying out the brackets).or you can use Pascal's triangle.
2008-04-16 4:04 pm
basically (2m-n)*(2m-n)*(2m-n)*(2m-n)
then use foil to (4m^2 - 4mn + n^2)*(4m^2 - 4mn + n^2)
then use foil to (16m^4 .....
I dont have the time to solve the rest
sorry
then use foil to (4m^2 - 4mn + n^2)*(4m^2 - 4mn + n^2)
then use foil to (16m^4 .....
I dont have the time to solve the rest
sorry
參考: MY BRAIN
2008-04-16 4:04 pm
(2m)^4-4C1(2m)^3 (n)+4C2 (2m)^2 (n)^2 - 4C3 (2m)^(1) (n)^3 + n^4
4C1=4
4C2=6
4C3=4
16m^4-8m^3n+24m^2n^2-8n^3+n^4
4C1=4
4C2=6
4C3=4
16m^4-8m^3n+24m^2n^2-8n^3+n^4
2008-04-16 4:02 pm
(2m - n)^4
= (2m - n)(2m - n)(2m - n)(2m - n)
= (4m^2 - 4mn + n^2)(2m - n)(2m - n)
= (8m^3 - 8m^2n + 2mn^2 - 4m^2n + 4mn^2 - n^3)(2m - n)
= 16m^4 - 16m^3n + 4m^2n^2 - 8m^3n + 8m^2n^2 - 2mn^3 - 8m^3n + 8m^2n^2 - 2mn^3 + 4m^2n^2 - 4mn^3 + n^4
= 16m^4 - 16m^3n - 8m^3n - 8m^3n + 4m^2n^2 + 8m^2n^2 + 8m^2n^2 + 4m^2n^2 - 2mn^3 - 2mn^3 - 4mn^3 + n^4
= 16m^4 - 32m^3n + 24m^2n^2 - 8mn^3 + n^4
= (2m - n)(2m - n)(2m - n)(2m - n)
= (4m^2 - 4mn + n^2)(2m - n)(2m - n)
= (8m^3 - 8m^2n + 2mn^2 - 4m^2n + 4mn^2 - n^3)(2m - n)
= 16m^4 - 16m^3n + 4m^2n^2 - 8m^3n + 8m^2n^2 - 2mn^3 - 8m^3n + 8m^2n^2 - 2mn^3 + 4m^2n^2 - 4mn^3 + n^4
= 16m^4 - 16m^3n - 8m^3n - 8m^3n + 4m^2n^2 + 8m^2n^2 + 8m^2n^2 + 4m^2n^2 - 2mn^3 - 2mn^3 - 4mn^3 + n^4
= 16m^4 - 32m^3n + 24m^2n^2 - 8mn^3 + n^4
2008-04-16 3:57 pm
u use a calculator ...
2008-04-16 3:56 pm
In order for something to be solved, it has to equal something.
What does (2m-n)^4 equal?
What does (2m-n)^4 equal?
收錄日期: 2021-05-01 10:22:04
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