F-(M+m）g=（M+m)a ····················☆2f-Mg=Ma·····································★联立上面两个星星,怎么得出!F= 〔2（m+M）〕/MF-(M+m）g=（M+m)a ···········

来源：学生作业帮助网 编辑：作业帮 时间：2024/11/06 00:47:09
F-(M+m）g=（M+m)a ····················☆2f-Mg=Ma·····································★联立上面两个星星,怎么得出!F= 〔2（m+M）〕/MF-(M+m）g=（M+m)a ···········
xU_OP*Y5+HNuUIfUU\)wQfY s`Y'EH"FruDsvi.isY^ȷr =JL

F-(M+m）g=（M+m)a ····················☆2f-Mg=Ma·····································★联立上面两个星星,怎么得出!F= 〔2（m+M）〕/MF-(M+m）g=（M+m)a ···········
F-(M+m）g=（M+m)a ····················☆
2f-Mg=Ma·····································★
联立上面两个星星,怎么得出!
F= 〔2（m+M）〕/M
F-(M+m）g=（M+m)a ····················☆
2f-Mg=Ma·····································★
联立上面两个星星，怎么得出！
F= 〔2f（m+M）〕/M

F-(M+m）g=（M+m)a ····················☆2f-Mg=Ma·····································★联立上面两个星星,怎么得出!F= 〔2（m+M）〕/MF-(M+m）g=（M+m)a ···········
(1)得F=(M+m)(g+a)
(2)得2f=M(g+a) (g+a)=2f/M
代入得
F=2f(M+m)/M

整体隔离法，第一个式子是整体分析，第二个式子隔离分析一个

F-(M+m）g=（M+m)a ····················☆
F=(M+m)(g+a)

2F-Mg=Ma·····································★
2F=M(g+a)
??题目抄错了？没错！请采纳好评

F-(M+m）g=（M+m)a ····················☆
F...

全部展开

F-(M+m）g=（M+m)a ····················☆
F=(M+m)(g+a)

2F-Mg=Ma·····································★
2F=M(g+a)
??题目抄错了？

收起

怎么有两个不一样F和f

观察有一个参数a

将a挪到一边

缺一个式子吧

很明显不能得出
f没法消掉。

·(m+an)/(m+bn)=b/a,求m F-(M+m）g=（M+m)a ····················☆2f-Mg=Ma·····································★联立上面两个星星,怎么得出!F= 〔2（m+M）〕/MF-(M+m）g=（M+m)a ··········· m-2m-3m+4m-5m-6m=m-2m-3m+4m-5m-6m+···+2002m-2003m-2004m 若(m+n)^2-mn(m+n)=(m+n)·M,则M是（） A.m^2+n^2 B.m^2-mn+n^2 C.m^2-3mn+n^2 D.m^2+mn+n^2 m^2+m-6/m^2+2m+1÷m+3/m^+m·m^2-1/2-m 其中m=-1/2 计算：m-m²/m²-1÷m/m-1·（m+1/m-1）² 如图所示,在光滑水平地面上,物体A和B的质量分别为m和M,在水平推力F作用下,物体A和B一起加速运动,两者保持相对静止,则A、B间的相互作用力为A.F B.M+m/M·F C.M/M+m·F D.m/M+m·F（·是乘的意思） 已知m·m+m-1=0,则m^3+2m^2+2004=? 根据幂的运算法则证明a^m·a^m=a^n+m 若（m+n)(m-n)^2-mn(m+n)=(m+n)·M,则M是 请大家发表下自己的解法,学过高数的进来看看函数f(x),g(x)在区间【a,b】 上连续可导,且导数均不为0,求证：存在一点m∈（a,b）,使得f(m)/f'(m) + g(m)/g'(m) =f(a)/f'(m) + g(b)/g'(m) 先化简,再求值：m-m²/m²-1÷m/m-1·(m+1/m-1)²,其中m=2. 物理··子弹打木块问题··由f=(Mmv0^2)/[2(M+M)d]怎么得出s2=md/(m+M）? 已知f（x）=x·eˆ-x,x∈[-2,2]的最大值为M,最小值为m,则M-m= 已知a>0 b>0 且m,n属于N 求证a^(m+n)+b^(m+n)>=a^m·b^n+a^n·b^m 设函数f(n)=lg[(1ⁿ+2ⁿ+3ⁿ+···+（m-1）ⁿ+mⁿa)/m],其中a∈R,m是给定的正整数,且m≥2.如果不等式f(x)>（x-1)lgm在区间【1,+∞）上有解,则实数a的取值范围是?答案是a>(3-m)/2,为什么? 若函数f(x)=log2x+x^2-2,g(x)=2^x+x-2若实数m,n满足f(m)=0,g(n)=0,则A,g(m)＜0＜f(n)B,0＜g(m)＜f(n）C,f(n)＜0＜g(m)D,f(n)＜g(m)＜0 计算2m+6/m²-4m+4·1/m+3·（m-2)(m+3)/m+3