若方程7x的平方-(k 13)x k的平方-k-2=0
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![若方程7x的平方-(k 13)x k的平方-k-2=0](/uploads/image/f/6976322-26-2.jpg?t=%E8%8B%A5%E6%96%B9%E7%A8%8B7x%E7%9A%84%E5%B9%B3%E6%96%B9-%28k+13%29x+k%E7%9A%84%E5%B9%B3%E6%96%B9-k-2%3D0)
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
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