若方程7x的平方-(k 13)x k的平方-k-2=0-搜答案-智慧作业帮

2/x²-7x+2=2/x²+x+6所以x²-7x+2=x²+x+68x=-4x=-1/2

(x的平方-X)分之一+(x的平方+x)分之K-5=(x的平方-1)分之K-1x+1+(k-5)(x-1)=x(k-1)x+1+kx-5x-k+5=kx-x-4x-k+6+x=0-3x-k+6=0k=

x(x^2-2)+2x^2(x+3)-3x(5x-7)=x(3x^2-9x+17)+2=>x^3-2x+2x^3+6x^2-15x^2+21x=3x^3-9x^2+17x+2=>2x=2x=1

(1)x^2-3x+√x^2-3x+5=7设x^2-3x为a则原式为a+√a+5=7解得a1=4,a2=11把a1=4,a2=11代入原式得：x^2-3x=4,x^2-3x=11解得x1=1,x2=2

1-x分之3-x+x的平方-8x+7分之x的平方-2=1+7-x分之5-x(x-3)/(x-1)+(x²-2)/[(x-7)(x-1)]=1+(x-5)/(x-7)平方同乘以(x-1)(x-

x³+3x²+3x-7=0x³-x²+4x²+3x-7=0x²(x-1)+(x-1)(4x+7)=0(x-1)(x²+4x+7)=

【解法一】【解法二】5x(x+7)(x+4)=5x(x^2+11X)5x(x²+11x+28)-5x(x²+11x)=05x^3+55x^2+140x=5x^3+55x^25x(x

解方程：(1)(x²-5x)/(x+1)+24(x+1)/x(x-5)+14=0；(2).2(x²+1)/(x+1)+6(x+1)/(x²+1)=7；(3)(x̾

您的题目好像有问题,是不是x²=7x-6解x²=7x-6x²-7x+6=0（x-1)(x-6)=0x1=1x2=6解x²=7x+6x²-7x-6=0△

2x²-7x=0x²-7/2+49/16=49/16（x-7/4）²=49/16x-7/4=+,—7/4x=0,x=7/8

7/X(X+1)+3/X(X-1)=6/(X+1)(X-1),方程两边都乘以X(X+1)(X-1),得：7(X-1)+3(X+1)=6X10X-4=6X4X=4X=1经检验X=1是增根,∴原方程无解.

7x^2+2x＝0x﹙7x+2﹚=0∴x=0或﹙7x+2﹚=0∴x1=0x2=-2／7

(3x-5)的平方-(3x+7)的平方=0[(3x-5)+(3x+7)][(3x-5)-(3x+7)]=0-12(6x+2)=0x=-1/3

是的

十字相乘

原方程可化为y+10/y=7y^2-7y+10=0(y-2)(y-5)=0y1=2,y2=5y1=x^2-3/x=2x^2-3=2xx^2-2x-3=0(x-3)(x+1)=0x1=3,x2=-1y2

X=1啊

x=1当m=__3__时,方程x-(x+2)分之m=（x+2)分之（x-1)会产生增根