2,5,7,9 .1,4 ,10,q .1,6,j,k .1,3,4,6.怎么算24点

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2,5,7,9 .1,4 ,10,q .1,6,j,k .1,3,4,6.怎么算24点

2,5,7,9 .1,4 ,10,q .1,6,j,k .1,3,4,6.怎么算24点
2,5,7,9 .1,4 ,10,q .1,6,j,k .1,3,4,6.怎么算24点

2,5,7,9 .1,4 ,10,q .1,6,j,k .1,3,4,6.怎么算24点
1: 5 × 7 -(2 + 9)
2: (5 × 7) - (2 + 9)
3: 5 × 7 - 2 - 9
4: (5 × 7) - 2 - 9
5: (5 × 7 - 2) - 9
6: ((5 × 7) - 2) - 9
7: 5 × 7 -(9 + 2)
8: (5 × 7) - (9 + 2)
9: 5 × 7 - 9 - 2
10: (5 × 7) - 9 - 2
11: (5 × 7 - 9) - 2
12: ((5 × 7) - 9) - 2
13: 7 × 5 -(2 + 9)
14: (7 × 5) - (2 + 9)
15: 7 × 5 - 2 - 9
16: (7 × 5) - 2 - 9
17: (7 × 5 - 2) - 9
18: ((7 × 5) - 2) - 9
19: 7 × 5 -(9 + 2)
20: (7 × 5) - (9 + 2)
21: 7 × 5 - 9 - 2
22: (7 × 5) - 9 - 2
23: (7 × 5 - 9) - 2
24: ((7 × 5) - 9) - 2
1: 4 ÷ (1 - 10 ÷ 12)
2: 4 ÷ (1 - (10 ÷ 12))
3: (4 × (10 - 1)) - 12
4: 4 × (10 - 1) - 12
5: (10 - 1) × 4 - 12
6: ((10 - 1) × 4) - 12
7: (10 ÷ (1 + 4)) × 12
8: 10 ÷ (1 + 4) × 12
9: 10 ÷ ((1 + 4) ÷ 12)
10: (10 ÷ (4 + 1)) × 12
11: 10 ÷ (4 + 1) × 12
12: 10 ÷ ((4 + 1) ÷ 12)
13: 10 × (12 ÷ (1 + 4))
14: 10 × 12 ÷(1 + 4)
15: (10 × 12) ÷ (1 + 4)
16: 10 × (12 ÷ (4 + 1))
17: 10 × 12 ÷(4 + 1)
18: (10 × 12) ÷ (4 + 1)
19: (12 ÷ (1 + 4)) × 10
20: 12 ÷ (1 + 4) × 10
21: 12 ÷ ((1 + 4) ÷ 10)
22: (12 ÷ (4 + 1)) × 10
23: 12 ÷ (4 + 1) × 10
24: 12 ÷ ((4 + 1) ÷ 10)
25: 12 × (10 ÷ (1 + 4))
26: 12 × 10 ÷(1 + 4)
27: (12 × 10) ÷ (1 + 4)
28: 12 × (10 ÷ (4 + 1))
29: 12 × 10 ÷(4 + 1)
30: (12 × 10) ÷ (4 + 1)
1: (1 + 11 × 13) ÷ 6
2: (1 + (11 × 13)) ÷ 6
3: (1 + 13 × 11) ÷ 6
4: (1 + (13 × 11)) ÷ 6
5: (11 × 13 + 1) ÷ 6
6: ((11 × 13) + 1) ÷ 6
7: (13 × 11 + 1) ÷ 6
8: ((13 × 11) + 1) ÷ 6
1: 6 ÷ (1 - 3 ÷ 4)
2: 6 ÷ (1 - (3 ÷ 4))

求和1+3q+5q^2+7q^3+9q^4= 高一数学 1+3q+5q*2+7q*3+9q*4=? 2q^9=q^3+q^6 即 2q^7=q+q^4 5q^4+q^3+q^2-3q+1=0求q 下列各组三个电荷可以平衡的是?A 4Q 4Q 4Q B 4Q -5Q 3Q C 9Q -4Q 36Q D -4Q 2Q -3Q 算24点[12题]7、5、7、6K、8、J、9Q、Q、4、4Q、J、K、Q5、7、A、10Q、6、9、107、4、Q、42、2、7、102、4、5、89、Q、4、75、K、6、7A=1 J=11 Q=12 K=13 q^5+q^4-2q^3-2q^2+1=0 matlab中两个函数想减,怎么办?函数里有三个未知数.f(p,q,x)-f(p-1,q+1,x),之后再化简下.f(p,q,x)=x^5 + (- 5*p - 5*q - 7)*x^4 + (9*p^2 + 19*p*q + 27*p + 9*q^2 + 27*q + 18)*x^3 + (- 7*p^3 - 25*p^2*q - 37*p^2 - 25*p*q^2 - 73*p*q - 48* 已知等比数列{an}的公笔q=-1/3,则a1+a3+a5+a7/a2+a4+a6+a8等于?为什么不是a1(1+2q+4q+6q)/a1(q+3q+5q+7q)=1+12q/16q=-3/4? 函数求利润的疑惑某厂的总收益函数和总成本函数分别为:R(Q)=18QC(Q)=Q^3-9Q^2+33Q+10 (Q为产量)求Q为何值时利润最大,并求出最大利润W=R(Q)-C(Q)=-Q^3+9Q^2-15Q-10dW/dQ=-3Q^2+18Q-15=0→Q(1)=1,Q(2)=5d^2W/dQ^2=-6Q+1 迅雷白金会员有的发8 7 2 9 5 4 8 9 1@q q 邮 怎么把(1-q^6)/(1-q^2)=7化简成1+q^2+q^4=7 无穷级数1-2q+3q^2-4q^3+5q^4-...的极限是什么? C语言多维数组问题main(){int a[5]={1 2 3 4 5}; b[4]={6 7 8 9}; *p=a *q=b; p+=2; q+=3;*p=*(q)+3;q--; *(q-1)=*(p+1)-5;printf(%d,%d,*p,*q); }求具体推算过程 尤其是解释一下*p=*(q)+3;*(q-1)=*(p+1)-5; 秋节四元一次方程式4x+2y+8z+16Q=682x-12y+12z+q=7x+13y-11z+11q=256x+11y-5z-10q=93x+4y-z+5q=632x+2y-3z+3q=283x-y+3z+3q=237x-6y-2z-q=1 q^5+q^6=2q^9怎么解 1 5 8 2 5 4 0 6 7 @ q q .c o m lim[(1+q)(1+q^2)(1+q^4)……(1+q^2^n)] (0