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    t趋于无穷,求(2^t+2^2t)/(1+2^2t)的极限?

    发布时间:2020-03-21

    2t³-6t²+6t-1=02t³-4t²+2t-2t²+4t-2+1=0
    2t(t²-2t+1)-2(t²-2t+1)+1=0
    2t(t-1)²-2(t-1)²+1=0
    2(t-1)³=-1
    (t-1)³=-1/2
    t-1=-³√0.5
    ∴t=1-³√0.5

    回复:

    y=(t-1)²/2+t
    =1/2 (t²-2t+1)+1/2×2t
    =1/2 (t²-2t+1+2t)
    =1/2(t²+1)
    =1/2t² +1/2

    回复:

    显然dx/dt=(t²)' *t² *lnt²=2t^3 *2lnt=4t^3 *lnt

    而dy/dt= -(t²)' *(t²)² *lnt²= -2t *t^4 *2lnt= -4t^5 *lnt

    所以得到
    dy/dx=(dy/dt)/(dx/dt)= -t²
    于是就解得
    d²y/dx²=d(dy/dx)/dt *dt/dx
    = d(-t²)/dt * 1/(4t^3 *lnt)
    = -2t /(4t^3 *lnt)
    = -1/(2t² *lnt)
    故d²y/dx²= -1/(2t² *lnt)

    回复:

    y'=3x²+2x 设切点为(t, t³+t²-1) 则切线方程为y=(3t²+2t)(x-t)+t³+t²-1=(3t²+2t)x-2t³-t²-1 1) 代入点m(1,2): (3t²+2t)-2t³-t²-1=2 化为:2t³-2t²-2t+3=0 求得t, 代入...

    回复:

    v=dx/dt=2-2t a=dv/dt=-2 t=0,v0=2 s=v0*t+0.5at^2=2x4-16=-8 应当先求得,何是往回走,即vt=0,t=1,余下的就......

    回复:

    令t=1, 2, ...k-1代入差分方程得: y(2)-y(1)=2 y(3)-y(2)=2*2² ... y(k)-y(k-1)=2(k-1)² 以上k-1个式子相加得: y(k)-y(1)=2[1²+2²+..+(k-1)²] =2(k-1)k(2k-1)/6 =k(k-1)(2k-1)/3 因此有y(k)=y(1)+k(k-1)(2k-1)/3

    回复:

    2t³-6t²+6t-1=02t³-4t²+2t-2t²+4t-2+1=0 2t(t²-2t+1)-2(t²-2t+1)+1=0 2t(t-1)²-2(t-1)²+1=0 2(t-1)³=-1 (t-1)³=-1/2 t-1=-³√0.5 ∴t=1-³√0.5

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