. 计算 (x2−4x2−4x 4 2−xx 2)÷xx−2 的结果是( )
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要使根号(x2+2x+4)-根号(x2-x+1)
∵x2-4x+y2+6y+z−3+13=0,∴(x-2)2+(y+3)2+z−3=0,∴x-2=0,y+3=0,z-3=0,解得x=2,y=-3,z=3,∴(xy)z=[2×(-3)]3=-216.
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
(1)原式=x(x+9)x(x+3)+(x+3)(x−3)(x+3)2=x+9x+3+x−3x+3=2(x+3)x+3=2;(2)原式=-x−2x−1÷x2−4x−1=-x−2x−1•x−1(x+2)
是二轮驱动,前面的“4”表示车轮总数,后面的“2”表示驱动轮数就象很多SUV上有4X4,就是总共四个车轮,四个车轮都是驱动轮.也有特种车,会有8X8,或6X6之类的
1/(x2-5x+6)-1/(4x-x2-3)-1/(3x-x2-2)=1/(x2-5x+6)+1/(x2-4x+3)+1/(x2-3x+2)=1/(x-2)(x-3)+1/(x-3)(x-1)+1/
原式=-x2+5+4x+5x-4+2x2=x2+9x+1.
2^5-5x2^4+10x2^3-10x2^2+5x2-1=(2-1)^5=1再问:有过程吗再答:晕,这是二项式的展开式啊,就这样:(a+b)^n=a^n+C(n,1)a^(n-1)b+C(n,2)a
∵−12≤x≤1,∴x-1≤0,x-3<0,2x+1≥0,∴x2−2x+1+x2−6x+9+4x2+4x+1=(x−1)2+(x−3)2+(2x+1)2=|x-1|+|x-3|+|2x+1|=1
∵x2-9=0,∴x=±3,当x=3时,x2-4x+3=0,∴x=3不满足条件.当x=-3时,x2-4x+3≠0,∴当x=-3时分式的值是0.故选C.
(1)f(x)=3sin(x+π2)+sinx=3cosx+sinx(2分)=2(12sinx+32cosx)=2sin(x+π3).(4分)所以f(x)的最小正周期为2π.(6分)(2)∵将f(x)
原式=-x+2x2+5+4x2-3-6x=6x2-7x+2.
设x2-3x=y,则原方程可化为:y+3y=4.即:y+3y−4=0.故选A.
12.5x(4x2.5)+(12.5x0.8)x2.5=125+25=150
原式=5x²-x²-(4x-x²)+2(x²-3x)=4x²-4x+x²+2x²-6x=7x²-10x
=0.3×(2.5×0.4)=0.3×1=0.3
1X2+2X3+3X4+4X5...+2013X2014==1/3(1×2×(3-0)+2×3×(4-1)+3×4×(5-2)+.+2013×2014×(2015-2012))=1/3×(1×2×3-
(5x²+3x-4)-(2x²+x+5)+(2x²-x+5)=5x²+3x-4-2x²-x-5+2x²-x+5=5x²+x-4
原式=2x/[(x-2)(x+1)]*(x+1)/(x-1)-x(x+2)/[(x+2)(x-2)]=2x/[(x-2)(x-1)]-x/(x-2)=[2x-x(x-1)]/[(x-2)(x-1)]=