5x 2>3[x-1],1 2≤8-3 2 2a4个整数解实数a取值范围
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![5x 2>3[x-1],1 2≤8-3 2 2a4个整数解实数a取值范围](/uploads/image/f/352135-55-5.jpg?t=5x+2%3E3%5Bx-1%5D%2C1+2%E2%89%A48-3+2+2a4%E4%B8%AA%E6%95%B4%E6%95%B0%E8%A7%A3%E5%AE%9E%E6%95%B0a%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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=
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