作业帮1 (1-x)-3 (1-x2)的极限
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 20:05:34
1/10x4+3x2+1=x4-x3+(x3+3x2+x)-x+1=x4-x3+x(x2+3x+1)-x+1=x4-x3-x+1=x4-(x3+3x2+x)+3x2+1=x4-x(x2+3x+1)+3
∵x2+x-1=0，∴x2+x=1，2x3+3x2-x=2x（x2+x）+x2-x=2x+x2-x=x2+x=1．
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&#
去分母得：x^2(y-1)+x(1-y)+y=0y=1时,上式无解y=1时,为二次式,须有delta>=0即(1-y)^2-4y(y-1)>=0(y-1)(3y+1)再问：x^2(y-1)+x(1-y
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
x^2-3x+2=0(x-2)(x-1)=0x=2或x=1当x=2时x^2+1/x^2=2^2+1/2^2=4+1/4=17/4当x=1时x^2+1/x^2=1^2+1/1^2=1+1=2
x+1/x=3(x+1/x)^2=3^2=9x^2+1/x^2+2=9x^2+1/x^2=7
原式=(x²+3x+9)/(x-3)(x²+3x+9)-6x/x(x-3)(x+3)-(x-1)/2(x+3)=1/(x-3)-6/(x-3)(x+3)-(x-1)/2(x+3)=
由条件左右除以x,得x+1/x=3平方得：x2+1/x2+2=9所以x2+1/x2—2=5所求值为根号5
x^2+3x+1=0方程两边同除以xx+3+1/x=0x+1/x=-3x^2+1/x^2=(x+1/x)^2-2=(-3)^2-2=9-2=7
x3-3x2-11x+8=x（x2-3x)-11x+8=x-11x+8=-10x+8
x/x²-3x+1=1/5x²-3x+1=5xx²+1=8xx+1/x=8平方x²+2+1/x²=64x²+1/x²=62x
答：x²-5x=3(x-1)(2x-1)-x(x+3)+15/(x²-3)=2x²-3x+1-x²-3x+15/(5x)=x²-5x-x+1+3/x=
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
x/(x²-3x+1)=2(x²-3x+1)/x=1/2翻过来x+1/x=7/2(x^4+x2+1)/x²翻过来=x²+2+1/x²=(x+1/x)&
原式=2x2-1，当x=-3时，原式=2×（-3）2-1=17．
对分母x^4+x^2+1进行因式分x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1),又因为x^2-3x+1=0,所以x^2=3x-1.
x2/x4+x2+1上下除以x^2=1/(x^2+1+1/x^2)=1/[(x+1/x)^2-1]=1/(9-1)=1/8
你好x²+3x-1=0两边同除以xx+3-1/x=0x-1/x=-3两边平方得x²+1/x²-2=9x²+1/x²=11x²+1/x
X2-3X-1=0则X-3-1/X=0则X-1/X=3则(X-1/X)²=3²=9则X²-2+1/X²=9则X²+1/X²=9+2=11