3x 2y=k-1 2x 3y=2k 中x与y的和为2求k的值
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不知道你学过二项式定理吗?知道组合数C(n,m)吗?假设你已经学过的话,看看下面的推导公式(n-1)^k=n^k+C(k,1)*n^(k-1)*(-1)+C(k,2)*n^(k-2)*(-1)^2+.
ans=-1/12*(324+12*633^(1/2))^(1/3)-2/(324+12*633^(1/2))^(1/3)-1/2;1/24*(324+12*633^(1/2))^(1/3)+1/(3
原式=2x2y+2xy-3x2y-3xy-4x2y=-5x2y-xy当x=-2,y=12时,原式=-9.
证明:K/(K+1)+1/[(K+1)(K+2)]=[K(K+2)+1]/[(K+1)(K+2)](注:通分,公分母为[(K+1)(K+2)])=(K+2K+1)/[(K+1)(K+2)]=(K+1)
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
本题需利用定积分求极限,其关键是构造1/n-->dx,i/n-->x,积分区间为x属于[0,1],于是分母提个n出来得:原式=(n-->+无穷)lim[(1^k+2^k+...+n^k)/(n^k)]
k^2-2k=3k^2-2k-3=0(k-3)(k+1)=0k不=3,则有k+1=0所以,K=-1
原式=x4+x3y+4x3y+x2y+4x2y2+4x2y2+xy2+4xy3+xy3+y4,=x3(x+y)+4x2y(x+y)+xy(x+y)+4xy2(x+y)+y3(x+y),=-x3-4x2
3×k×k-2k-1=-13k^2-2k=0k(3k-2)=0得k-0或k=2/3
6k^2-15k+3k-6k^2=36则-12k=36k=-3再问:(-4x-y)(-5x+2y)=(3/4×a的n+1次方-1/2×b)×2ab=
原式=(x^4-2x²y²+y^4)+6xy(x²+2xy+y²)-2xy(x+y)=(x²-y²)²+6xy(x+y)²
两实1.652310139-0.431990495两虚-0.61015986+1.957209i-0.61015986-1.957209i
(3k-2)(2+k)-(2k+1)(3-2k)=0(3k²+4k-4)-(-4k²+4k+3)=03k²+4k-4+4k²-4k-3=07k²-7=
3k(2k-5)+2k(1-3k)=526k²-15k+2k-6k²=52-13k=52k=-4当k等于-4时,3k(2k-5)+2k(1-3k)=52再问:3Q,你QQ是多少,我
原式=[x3y2-x2y-x2y+x3y2]÷3x2y=(2x3y2-2x2y)÷3x2y=23xy-23;当x=3,y=-1时,原式=23×3×(-1)-23=-83.
13/12(p-q)
k²+k-1=0,得k²=1-kk²+k=1k³+2k²+2009=(k+2)k²+2009=(k+2)(1-k)+2009=2-(k
x4-xy3-x3y-3x2y+3xy2+y4=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-
先简化一下题目:5K+K/3+1=2001>>>>(16/3)K=2001-1>>>16K=2000*3>>>>K=6000/16>>>K=375
2x+3y=-k+2,①3x-2y=5k+3②2*①+3*②13x=13k+13所以x=k+1代入①y=-kx-y=2k+1=5k=2