1-(1/4)t=2[1-(1/3)t]怎么解

(2t+3)(2t-)-(4t+1)(t-9) 4t^-8t+5-3t^+6t-2 其中t=2/1 (t^5+t^4+t^3+t^2+t+1)(t-1) 用直式计算 画出下列信号的波形1）f(t)=tu(t); 2)f(t)=t[u(t)-u(t-2)]; 3)f(t)=(t-1)u(t-1); 4)f(t)=sint/t画出图形~ 已知f(t)=3t*t-2t-2/t+3/t*t,证明f(t)=f(1/t) 5t-3≤4t-1 7+2t>6+3t 3+t 5t-3≤4t-1 7+2t>6+3t 3+t 4t^3-3t^2-2t+1=0 t等于多少 [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),'.') 化简t+1/t-(t+2)²-1/t²×t+1/t+3 3t^2+3t-1-t^3=0, matlab中怎么对下面的程序进行二维插值并画出三维视图呢?[x,y]=meshgrid(-2:1:2);t=2;Z=[(sin(t))^3 t^2 sin(t) cos(t) (sin(t))^2;t^2 sin(t) (cos(t))^4 cos(t) (sin(t))^2;t sin(t) cos(t) t^2 (sin(t))^2;sin(t) (cos(t))^3 t^2 (sin(t))^2 79.788456t - 1250t^2 - 8311.297508t^3 = 1 当t=1时,代数式t^3-2t[2t^2-3t(2t+2)]的值 3t^2-1=6t 利用行列式的性质证明| 1 1 1 || t t^2 t^3 |=t^4(t-1)^3(t+1) [这里的数字都是次方)| t^2 t^4 t^6 | 用解析法求下列二阶微分方程(1) y(t) + 4y(t) + 3y(t) = f (t),f (t) = ε (t)(2) y(t) + 4y(t) + 4y(t) = f '(t) + 3 f (t),f (t) = e−tε (t)注：e-t为e的-t次方其中ε (t)为阶跃函数,即当t0时,ε (t)=1 t/(t^2-t+1)的积分