2x(½x2-1)(
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/24 04:17:02
要使根号(x2+2x+4)-根号(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
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
2/(x2-x)+6/(1-x2)=7/(x2+x)2/x(x-1)-6/(x-1)(x+1)=7/x(x+1)[x*(x-1)*(x+1)]*[2/x(x-1)-6/(x-1)(x+1)]=[7/x
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)=
x3+x2=x2(x+1),x2+2x+1=(x+1)2,x2-x-2=(x+1)(x-2),∴它们的公因式为x+1.故答案为:x+1.
X^2+2x-1=0x=[-2±√(2^2+4)]/2=-1±√2X^2+2x-1=(x+1-√2)(x+1+√2)
原式=(x+1)/(x-1)-x(x-2)/(x+1)(x-1)÷(x-2)(x+1)/(x+1)²=(x+1)/(x-1)-x/(x-1)=(x+1-x)/(x-1)=1/(x-1)请好评
要过程吗?再问:要再答:
(x²-1)/(x²-2x+1)÷(x²+x)/(x-1)=(x+1)(x-1)/(x-1)²乘(x-1)/x(x+1)=1/x
原式=x3+8+x3-1=2x3+7=-16/27+7=173/7
x2+x+1=2/(x2+x)(X²+x)²+(x²+x)-2=0(x²+x+2)(x²+x-1)=0∴x²+x-1=0x=(-1±√5)/
两边乘x(x+1)(x-1)2(x-1)+3(x+1)=4x2x-2+3x+3=4x5x+1=4xx=-1经检验,x=-1时分母x+1=0增根,舍去方程无解
(x²+x)(x²+x-2)=-1把(x²+x)看成整体(x²+x)[(x²+x)-2]=-1运用乘法分配率(x²+x)²-2(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
原式=-2x2+3x-5x+2x2+1+x2=x2-2x+1.
令a=x+1/xa²=x²+2+1/x²2(a²-2)-9a+14=0(2a-5)(a-2)=0x+1/x=5/22x²-5x+2=0(2x-1)(x
把题拍过来帮你解
解题思路:这个是因式分解问题。由完全平方公式,再应用换元法可以得到结果.解题过程:
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)]=