计算(b-a)(b-a)^3(a- b)^5 x^m×(x^n)^3÷(x^m-1×2x^n-1) (2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^51.(b-a)(b-a)^3(a- b)^5 2 .x^m×(x^n)^3÷(x^m-1×2x^n-1) 3.(2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5
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![计算(b-a)(b-a)^3(a- b)^5 x^m×(x^n)^3÷(x^m-1×2x^n-1) (2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^51.(b-a)(b-a)^3(a- b)^5 2 .x^m×(x^n)^3÷(x^m-1×2x^n-1) 3.(2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5](/uploads/image/z/1104075-27-5.jpg?t=%E8%AE%A1%E7%AE%97%28b-a%29%28b-a%29%5E3%28a-+b%29%5E5+x%5Em%C3%97%28x%5En%29%5E3%C3%B7%28x%5Em-1%C3%972x%5En-1%29+%282-6%29%5Em%2B3%C3%97%EF%BC%88b-a%29%5E2%C3%97%28a-b%29%5Em%C3%97%EF%BC%88b-a%EF%BC%89%5E51.%28b-a%29%28b-a%29%5E3%28a-+b%29%5E5+2+.x%5Em%C3%97%28x%5En%29%5E3%C3%B7%28x%5Em-1%C3%972x%5En-1%29+3.%282-6%29%5Em%2B3%C3%97%EF%BC%88b-a%29%5E2%C3%97%28a-b%29%5Em%C3%97%EF%BC%88b-a%EF%BC%89%5E5)
计算(b-a)(b-a)^3(a- b)^5 x^m×(x^n)^3÷(x^m-1×2x^n-1) (2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^51.(b-a)(b-a)^3(a- b)^5 2 .x^m×(x^n)^3÷(x^m-1×2x^n-1) 3.(2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5
计算(b-a)(b-a)^3(a- b)^5 x^m×(x^n)^3÷(x^m-1×2x^n-1) (2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5
1.(b-a)(b-a)^3(a- b)^5 2 .x^m×(x^n)^3÷(x^m-1×2x^n-1)
3.(2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5
计算(b-a)(b-a)^3(a- b)^5 x^m×(x^n)^3÷(x^m-1×2x^n-1) (2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^51.(b-a)(b-a)^3(a- b)^5 2 .x^m×(x^n)^3÷(x^m-1×2x^n-1) 3.(2-6)^m+3×(b-a)^2×(a-b)^m×(b-a)^5
1、因为b-a=-(a-b)所以(b-a)(b-a)^3=(a-b)^4答案是(a-b)^9
2、x^m÷x^m-1=x,(x^n)^3÷2x^n-1=1/2x^(3n-n+1)=1/2x^(2n+1),答案1/2x^(2n+2)
3、看不懂2-6什么意思,是不是打错了,我只能算到(2-6)^m+3×-(a-b)^m+7
我学这东西已经好多年的,可能有点变化,答案仅供参考
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