5x 2>3(x-1)

2x2+1-3x+7-3x2+5x=（2-3）x2+（-3+5）x+8，=-x2+2x+8．

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

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)=

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)2x2-5x+x2+4x,其中x=-3=3x²-x=3x(-3)²+3=27+3=30(2)(3x2-xy-2y2)-2(x2+xy-2y2),其中x=6,y=-1=3x&

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

方程两边同时乘以x(x+1)(x-1)得：5(x-1)+3(x+1)=7x解得：x=2检验：x(x+1)(x-1)=6所以x=2是原分式方程的解

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/

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

5x(x2-3x-1)-x2(1-x)=5x^3-15x^2-5x-x^2+x^3=6x^3-16x^2-5x

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=

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/

3x^2+15x-2+2(x^2+5x+1)^1/2=03(x^2+5x+1)+2(x^2+5x+1)^1/2-5=0另(x^2+5x+1)^1/2=y3y^2+2y-5=0y1=1y2=-5/3(舍

原式=5x²-x²-(4x-x²)+2(x²-3x)=4x²-4x+x²+2x²-6x=7x²-10x

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

已知X1X2为方程5X平方-3X-1=0两个根；所以x1+x2=3/5;x1x2=-1/5;x1-x2=√(x1-x2)²=√[(x1+x2)²-4x1x2]=√(9/25+4/5