方程x^2-5x=0,的解-搜答案-智慧作业帮

当x>=1时,x方-2x-(x-1)=5,(x-4)(x+1)=0,x=4当x

4/(x-2)+(x-1)/[(x-2)(x-3)]-2/(x-3)=0[4(x-3)+(x-1)-2(x-2)]/[(x-2)(x-3)]=03(x-3)/[(x-2)(x-3)]=0不知道解了~~

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

【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

只有一个实根.设f(x)=x^3+2x-19为单调增函数.所以只有一个实根.下面来求这个实根由于f(2)=-7f(3)=11所以这个根在（2,3）内.利用二分法求这个解.取x0=5/2f(5/2)=1

x的5次方-x的3次方-2x=0x(x^4-x^2-2)=0x(x^2+1)(x^2-2)=0x(x^2+1)(x+根号2)(x-根号2)=0x=0或x=-根号2或x=根号2再问：谢谢，请问还有一题怎

intf(floatx)返回值,错了.

2x^2-5x-x^2-2x=x^2-6-7x=-6x=6/7

两边除以5x²+x/5=1x²+x/5+1/100=1+1/100(x+1/10)²=101/100x+1/10=±√101/10x=(-1-√101)/10,x=(-1

x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)x/(x-2)=2x/(x-3)+(1-x)/(x-2)(x-3)x(x-3)/(x-2)(x-3)=2x(x-2)/(x-2)(x-3)

3（x-2）=5x（2-x）3（x-2）-5x（2-x）=03（x-2）+5x(x-2)=0(x-2)(3+5x)=0解得x=2或-3/54x（3x+8）-5（3x+8）=0(3x+8)(4x-5)=

3x^2-x(2x+3)=x(x-5)-2(2x+3)3x^2-2x^2-3x=x^2-5x-4x-66x=-6x=-1

（x+1)/(x+2)+(x+6)/(x+7)=(x+2)/(x+3)+(x+5)/(x+6)(x+2-1)/(x+2)+(x+7-1)/(x+7)=(x+3-1)/(x+3)+(x+6-1)/(x+

X-5=X/2 解方程

两边*22x-10=xx-10=0x=10

显然x=0是一个解x≠0两边除以xx|x|-5|x|+2=0x0,且x1+x2=5>0x1x2=2>0所以是两个正跟所以一共4个

3x的二次+5x=0x(3x+5)=0x1=0x2=-5/39x的二次=（x-2）的二次9x²-（x-2）²=0（3x+x-2）（3x-x+2）=0(4x-2)(2x+2)=0(2

(x的平方-2x)-5(x-2)=x(x-2)-5(x-2)=(x-2)(x-5)=0x-2=0或x-5=0x1=2x2=5

6/(x²+5x)-3/(2x+10)=06/(x²+5x)=3/(2x+10)2/x(x+5)=1/2(x+5)2/x=1/2x=4经检验：x=4是原方程的解.再问：确定对吗？再