计算:{[-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/08 13:44:33
计算:{[-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)]}/[-1/12x^(2n-1)y^(2n-1)]
(^*__*^)
(^*__*^)
(提公因式)
-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)
=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]
所以
原式=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]/[-1/12x^(2n-1)y^(2n-1)]
=2(x^2y^2-2x^2-3)
-1/6x^(2n+1)y^(2n+1)+1/3x^(2n+1)y(^(2n-1)+1/2x^(2n-1)y^(2n-1)
=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]
所以
原式=-1/6x^(2n-1)y^(2n-1)[x^2y^2+2x^2+3x^(2n-1)]/[-1/12x^(2n-1)y^(2n-1)]
=2(x^2y^2-2x^2-3)
信号与系统中 y(n)=x(n)+x(n+1)如果|x(n)|≤M, 则|y(n)|≤|x(n)|+|x(n+1)|≤2
分解因式:-2X^(5n+1)Y^n+4X^(3n+1)Y^(n+2)-2X^(n+1)Y^(n+4)
计算(-3x^n y)^2乘以2x^(n-1)y的结果是
y[n]=x[n]+0.5x[n-1]+0.7y[n-1]-0.1y[n-2] 怎么用matlab求y[n]的方程?或者
(1)[(x+y)^2n]^4除以(-x-y)^2n+1(n是正整数)
单项式除以单项式 -x^n-2y^n+1÷(-4x^n-4y^n-3)
若1+2+……+n=k,试探求:(x^n*y)*(x^(n-1)*y^2)*(x^(n-2)*y^3)*……*(x*y^
若1+2+3+...+n=a,求代数式(x^ny)(x^n-1y^2)(x^n-2y^3)...(x^2y^n-1)(x
计算(x^(2n)+x^n+1)(x^(3n)-x^(2n)+1)
已知实数x,y满足x+y=1,n∈N+,求证x^2n+y^2n≥1/2^(2n-1)
计算(1)(3m-4n)(3m+4n)-(2m-2n)(2m+n) (2)(x+5y)²-(x-5y)
用列举法表示 集合{x|x=(-1)^n,n∈N}{y|y=-x^2+6,x∈N,y∈N}{(x,y)|y=-x^2+6