(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简
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(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简(1+1/y)^(4/5)*y^(4/5)如何用ma
(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简
(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简
(1+1/y)^(4/5)*y^(4/5)如何用mathematica化简
In[1]=FullSimplify[(1 + y^(-1))^(4/5)*y^(4/5), y > 0]
Out[1]=(1 + y)^(4/5)
2(y-4)(3y+2)-5(-3y+7)(y+1),其中y=-1
1/2y+1=4y-2/5-y
y-1/y-2 - y-3/y-4=y-2/y-3 - y-4/y-5一共有四个分数哦~希望可以快一点~谢谢!
计算:[(y+1)/(y^2-4y+3)减去(y-2)/(y^2-6y+9)]除以(y-5)/(y-1)
解3y+5/3y-6=1/2+5y-4/2y-4
4(2y+5)-3y=7(y-5)+4(2y+1)
4(5—2y)+7(y+1)=3(4y-1)+17求y
4*(5—2y)+7(y+1)=3(4y-1)+17求y
代数式4y² - 2y + 5 = 7,那么代数式2y² - y + 1
计算(3y-1)(2y+3)-(6y-5)(y-4)
(3y-1)(2y-3)+(6y-5)(y-4)=?xie
(3y-1)(2y一3)十(6y一5)(y一4)
(3y-1)(2y-3)+(6y+5)(y-4)=?
若 4y^2-2y+5=7 则 2y^2-y+1=?
计算:(-3y)^2(2y+1)-(3y-5)(3y+4)
已知4y-1与5y-2y互为相反数,求y的值
w=(d 2*m 3*(m 1)/5 y y/4-y/100 y/400
[T,Y]=ode45(@rigid,[0 1],[rand(15,1)]);plot(T,Y(:,1),'-',T,Y(:,2),'-.',T,Y(:,3),'.',T,Y(:,4),'-',T,Y(:,5),'-.',T,Y(:,6),'.',T,Y(:,7),'-',T,Y(:,8),'-.',T,Y(:,9),'.',T,Y(:,10),'-',T,Y(:,11),'-.',T,Y(:,12),'-.',T,Y(:,13),'.',T,Y(:,14),'-',T,Y(:,15),'.')