为什么7x?6x=x不是是方程
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/18 00:54:45
![为什么7x?6x=x不是是方程](/uploads/image/f/1466434-10-4.jpg?t=%E4%B8%BA%E4%BB%80%E4%B9%887x%3F6x%3Dx%E4%B8%8D%E6%98%AF%E6%98%AF%E6%96%B9%E7%A8%8B)
(x+3)/(x+2)+(x+9)/(x+8)=(x+5)/(x+4)+(x+7)/(x+6)1+(x+5)/(x+2)(x+8)=1+(x+5)/(x+4)(x+6)x=-5再问:第二步那个1是怎么
将x-7/x-9分解成1+2/(x-9)其他分式同理则原方程等价于1/x-9+1/x-5=1/x-6+1/x-81/x-6-1/x-5=1/x-9-1/x-81/(x-5)(x-6)=1/(x-8)(
((X-2)(X-4)-(X-3)^2)/(X-3)(X-4)=((X-5)(X-7)-(X-6)^2)/(X-6)(X-7),-1/(X^2-7X+12)=-1/(X^2-13X+42),6X=30
[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)(x-4+x-8)/(x-4)(x-8)=(
42x-6x²=36-6x²+10x42x-10x=3632x=36x=9/8再问:解方程:(x+1)(x+4)=x的2次方-6再答:x²+5x+4=x²-65
x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/
我给个思路吧(X+2-1/X+2)-(X+3-1/X+3)=(X+6-1/X+6)-(X+7-1/X+7)1-1/(X+2)-1+1/(X+3)=1-1/(X+6)-1+1/(X+7)剩下的就自己解决
是的因为有未知数是等式再答:谢谢
/>因为:x-5/x-6+x-8/x-9=x-7/x-8+x-6/x-7,所以:1+1/(x-6)+1+1/(x-9)=1+1/(x-8)+1+1/(x-7)即:1/(x-6)+1/(x-9)=1/(
(1)(X+7)/(X+6)+(X+9)/(X+8)=(X+10)/(X+9)+(X+6)/(X+5)这类方程重在运算的技巧性,观察分母7+9=10+6;6+8=5+9所以先移项:(X+7)/(X+6
(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/
x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/
对含有未知数的等式叫方程,所以2x=10是方程x=0是方程的解,x不再是未知的了.x=0不是方程
明显少了一个条件x不等于0分式方程成立的条件之一就是分母不为0
通分化简得到(x+1)(x+2)=(x+5)(x+6)3x+2=11x+308x=-28x=-7/2
以上三等式都是一元一次方程,第一、三方程无解,第二方程的解为任意实数.方程中有m个未知数,这个方程就称为m‘元’方程;方程中次数最高的那个项为n次,那这个方程就称为n次方程,两者合起来就叫m元n次方程
是方程,解是所有数,包括实数虚数
x^6+x^5(-1-3)+x^4(-4+3+1)+x^3(7+12-1-2)+x^2(-21-4+2)+x(7+8)-14+6x^2-15x+18=0x^6-4x^5+16x^3-17x^2+4=0
x+62/x+63+x+68/x+69=x+63/x+64+x+67/x+68可以把x+62看做x解得x=-4所以x+62=-4x=-66
X-1/X-2-(X-2)/(X-3)=(X-5)/(X-6)-(X-6)/(X-7)(X-1)(X-3)-(X-2)²/(X-2)(x-3)=(X-5)(X-7)-(X-6)²/