5x 2>3[x-1],1 2≤8-3 2 2a4个整数解实数a取值范围-搜答案-智慧作业帮

原式=-x-2，当x=12时，原式=-12-2=-212．

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

把x2+5x看成是一个未知数Y,就可以解了

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

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

如图所示,条件区间为途中阴影部分.Z=x1+3x2的斜率=-1/3,Z为函数与Y轴交点的纵坐标.由图可知,当函数过点A时Z最大,求的A坐标为（2,4）,代入Z=x1+3x2得Z=14所以最大值为14有

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/

∵A=5x2+4x-1，B=-x2-3x+3，C=8-7x-6x2，∴A-B+C=（5x2+4x-1）-（-x2-3x+3）+（8-7x-6x2）=5x2+4x-1+x2+3x-3+8-7x-6x2=

x2-5x+1=0则x2+x2/1

你可以参见“韦达定理”方程两个根的积是1,说明他们互为倒数.x^2+1/x^2=(x+1/x)^2-2*x*1/x=（-5）²-2=23

(x2-x)(x2-x-8)+12解原式=（x²-x)(x²-x-8)+12=(x²-x)²-8(x²-x)+12=(x²-x-6)(x&#

直接写答案?再问：要写过程，再答：好了哦，给采纳一下吧

x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x

=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/

5x²－3x－5＝0△＝3²－4×5×(－5)＝109x＝[﹣(﹣3)±√109]/5由原方程可得所求式子＝(x＋5)－1/(x＋5)所求式子＝(118±6√109)/25－25/

原式=x3+5x2+4x-1+x2+3x-2x3+3+8-7x-6x2+x3=10，故与x无关．

原式=-2x2+3x-5x+2x2+1+x2=x2-2x+1．

X和*（乘号）无法区分,请详细点再问：没有乘号，X2就是X的平方再答：X2-4X-5≤0解(x+5)(x-1)=0x+5=0或x-1=0x=-5或x=1-2X2+3X-2＜0解2x^2+3x-2=0(

(x+1)²-2(x²-1)+(x-1)²=(x+1)²-2(x+1)(x-1)+(x-1)²=[(x+1)-(x-1)]²=2²