2(x 2)(x-1)-(x-3)(x 3)=(x-5)^2
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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
原式=x³-6x²-9x-2x²+12x+18-(x²-5x)(x-3)=x³-8x²+3x+18-(x³-3x²-5x
原式=3x2-x3+x3-2x2+1=x2+1=3+1=4
(x²+1)²-4x(x²-1)=(x²-1)²-4x(x²-1)+4x²=(x²-1-2x)²(x^4-2x
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
so easy、、提示一下,望采纳
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)=
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
要过程吗?再问:要再答:
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=
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
原式=[x+2x(x-2)-x-1(x-2)2]÷x2-16x2+4x=[x2-4x(x-2)2-x2-xx(x-2)2]÷x2-16x2+4x=x-4x(x-2)2•x(x+4)(x+4)(x-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
因为X*根号(X^2+3X+18)-X*根号(X^2-6X+18)=1则X*根号(X^2+3X+18)=X*根号(X^2-6X+18)+1两边平方得X^2*(X^2+3X+18)=1+X^2*(X^2
通分:(x^2-3x)+(2x-1)(x+1)=0化简:3x^2-2x-1=0x1=1(舍去,分母不为0)x2=-1/3
发照片呀再问:请原谅手机不行再问:请帮我一下,谢谢。再答:看不明白再问:你写出来就行了的再答: 再答:是这样吗再问:嗯,麻烦把过程写详细点再答:我说题目对不再问:是的再问:对的再问:还有一个
X+3X+9/X-27+6X/9X-X-X-1/6+2X=X-X+3X-X+2X+9/X-27+2/3-1/6=4X+9/X-27+(4-1)/6=4X+9/X-27+1/2=4X+9/X-(54-1
原式=2x2-1,当x=-3时,原式=2×(-3)2-1=17.
原式=-2x2+3x-5x+2x2+1+x2=x2-2x+1.
(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