1x2-3x4 5x6-7x8 9等于多少
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![1x2-3x4 5x6-7x8 9等于多少](/uploads/image/f/111910-22-0.jpg?t=1x2-3x4+5x6-7x8+9%E7%AD%89%E4%BA%8E%E5%A4%9A%E5%B0%91)
2x2+1-3x+7-3x2+5x=(2-3)x2+(-3+5)x+8,=-x2+2x+8.
设(x²-1)/(x²+2x)=t则8t+3/t=118t²-11t+3=0(8t-3)(t-1)=0解得t=3/8或t=11.t=3/8(x²-1)/(x
后面的x²+11x-708有误吧!再问:没有题目就这样能不能帮我再答:那我就试试:原式为:1/x2+x+1/x2+3x+2+1/x2+5x+6+1/x2+7x+12+1/x2+9x+20=5
x2+3x2+1=0中的3x2表示什么?再问:已知X2+3X+1=0,求X2+1/X2的值?得数是7。求过程?我打的是X的平方。怎么会出X2、再答:答:因为x≠0,两边都除以x得:x+1/x=-3,两
x=(4-√7)/3x²=(16+7-8√7)/9=(32-8√7)/9-9/9所以x²=(8/3)x-1即x²+1=(8/3)x所以x+1/x=8/3(x+1/x)
原式=[1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)]×(x+4)(x+5)=[1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x
3/x2=1/x2-x即3*2-3x=x*22x*2=3Xx=0(舍去)x=3/2
1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)同理1/(x²+5x+6)=1/(x+2)-1/(x+3)1/(x²
解题思路:吸纳化简,根据已知条件,整体代入可解。解题过程:
方程两边同时乘以x(x+1)(x-1)得:5(x-1)+3(x+1)=7x解得:x=2检验:x(x+1)(x-1)=6所以x=2是原分式方程的解
把x^2+3x看做为a(a+3)(a-7)+25=a^2-4a+4=(a-2)^2=(x^2+3x-a)^2
(x2+3/根号x2+1)^2-(2根号2)^2=(x^4-2x^2+1)/8(x^2+1)=(x^2-1)/8(x^2+1)>=0,又因为不等式两边均为正,所以x2+3/根号x2+1≥2根号2
7/(x+x2)-3/(x-x2)=6/(x2-1)两边同乘以x(x+1)(x-1),得7(x-1)+3(x+1)=6x7x-7+3x+3=6x10x-6x=3-74x=-4x=-1经检验x=-1是增
=1/(x-2)(x-1)+1/(x-2)(x-3)+1/(x-3)(x-4)-4/(x-1)(x-5)帮你删除了1/x2-7=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x
1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)=1/x-1/
原式=3x2-[7x-4x+3-2x2]=3x2-7x+4x-3+2x2=5x2-3x-3
令t=x^2+1>=1则x^2=t-1代入函数得:f=[(t-1)^2+3(t-1)+6]/t=[t^2-2t+1+3t-3+6]/t=[t^2+t+4]/t=t+4/t+1t+4/t>=2√(t*4
x²+x-1/(x²+x)=3/2两边同时乘以(x²+x)得:(x²+x)²-1=3(x²+x)/22(x²+x)²-3
102x89=(100+2)×89=100×89+2×89=8900+178=9078
如果是12的话就没有办法分解了,对不起(X^2+3x+2)(x^2+7x+12)-120=(x+1)(x+2)(x+3)(x+4)-120=(X^2+5x+4)(x^2+5x+6)-120=(X^2+