ln(x 1)-x3 3-x2 2
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99.99×22.22+33.33×33.34=33.33×3×22.22+33.33×33.34=33.33×66.66+33.33×33.34=33.33×(66.66+33.34)=33.33×
设平均数为a,方差公式展开可知,-2a(x1+x2+x3+x4+x5)+5a2=-2a.5a+5a2=-5a2,所以-5a2=-20,得到a=2,正确答案选B
∵函数f(x)=x33+12ax2+2bx+c∴f′(x)=x2+ax+2b=0的两个根为x1,x2,∵x1,x2分别在区间(0,1)与(1,2)内∴f′(0)>0f′(2)>0f′(1)<0⇒b>0
如果单单是目标函数错误,那么就是末尾缺少一个分号.如果不止,请贴出完整代码可以Hi联系我!Max=12*x31+7*x32+13*x33-0.5*x21-0.5*x22-0.5*x23-x41*x11
由△=36-4k≥0得k≤9,∵x12x22-x1-x2=115,x12x22-(x1+x2)=115,k2-6=115,k2=121,解得k=-11,或k=11(不合题意舍去),得x12+x22=(
应该是“x21+x22+…+x40的最大值为A”吧?如果是这样,因为这些变量都是正整数,当“x21+x22+…+x40”取最大时,“x1+x2+…+x20”中各个变量的值均为1,则A=38;同理,当“
(I)因为函数y=f(x)在[3,+∞)上为增函数,所以f′(x)=x[2ax2+(1−4a)x−(4a2+2)]2ax+1≥0在[3,+∞)上恒成立当a=0时,f′(x)=x(x-2)≥0在[3,+
75185.294052再答:👌✌🙏再问:要过程再问:要过程再答:不知道
∵x1,x2是方程mx2+2x+m=0的两个根∴x1+x2=-2/mx1x2=1△=4-4m²≥0,即-1≤m≤1但m≠0∴x1²+x2²=(x1+x2)²-2
有根,判别式大于等于016m²-16(m+2)>=0m²-m-2=(m+1)(m-2)>=0m=2x1+x2=mx1x2=(m+2)/4x1²+x2²=(x1+
方程x^2-x-1=0的两根为x1,x2,∴x1+x2=1,x1x2=-1.∴1/x1^2+1/x2^2=(x1^2+x2^2)/(x1x2)^2=(x1+x2)^2-2x1x2=1+2=3.
99.99×22.22+33.33×33.34=33.33×3×22.22+33.33×33.34=33.33×66.66+33.33×33.34=33.33×(66.66+33.34)=33.33×
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设有p个x取1,q个x取-2,有p−2q=−17p+4q=37,(5分)解得p=1q=9,(5分)所以原式=1×13+9×(-2)3=-71.(3分)
99.99×22.22+33.33×33.34=(99.99÷3)×(22.22×3)+33.33×33.34=33.33×66.66+33.33×33.34=33.33×(66.66+33.34)=
按你写这个我也看不懂.大概猜一下是(sin^2x+cos^2x)/(sinxcosx)*cos^2x=1/(sinxcosx)*cos^2x=cosx/sinx
99.99×22.22+33.33×33.34=(99.99÷3)×(22.22×3)+33.33×33.34=33.33×66.66+33.33×33.34=33.33×(66.66+33.34)=
∵方程x2-6x+2=0的两根之积为2,两根之和为6,∴x2x1+x1x2=x21+x22x1x2=(x1 +x2 )2−2x1x2x1x2=62−2×22=16.故答案为16.