分式x的平方减去2x加上m分之1
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/02 01:42:46
![分式x的平方减去2x加上m分之1](/uploads/image/f/2333496-48-6.jpg?t=%E5%88%86%E5%BC%8Fx%E7%9A%84%E5%B9%B3%E6%96%B9%E5%87%8F%E5%8E%BB2x%E5%8A%A0%E4%B8%8Am%E5%88%86%E4%B9%8B1)
原式1/(x-1)+(x^2-3x)/(x+1)(x-1),然后通分得(x+1+x^2-3x)/(x+1)(x-1)=(x^2-2x+1)/(x+1)(x-1)=(x-1)(x-1)/(x-1)(x+
能把题目说的更明白点么?
题目给错了吧,应该为x的平方加上2x加上y的平方减去6y加上10等于0如果是,x^2+2x+y^2-6y+10=0,所以:(x+1)^2+(y-3)^2=0所以:x+1=0且y-3=0两式相加,得到:
负2.根据题意,分子为零.解出X等于3或者-2,又因为分母不为零,即X≠3.所以X=-2.记得采纳啊
2x^2-2x+1/2=2(x^2-x)+1/2=2[(x^2-x+1/4)-1/4]+1/2=2(x-1/2)^2-1/2+1/2=2(x-1/2)^2采用的是配方法
答案是x+1-√2x
6+2m=0,m=-3
x+1的平方分之(x+1)(x-1)减去x-1分之x+1=(X-1)/(X+1)-(X+1)/(X-1)=[(X-1)^2-(X+1)^2]/[(X+1)(X-1)]=-4X/[(X+1)(X-1)]
解题思路:原分式方程化为整式方程后,将分式方程分式方程的增根代入,求出m解题过程:
答案:(3X-6)分之10减(3X+3)分之1
x/(x+3)-(x+1)/(x-2)=(x-2m)/(x^2+x-6)x/(x+3)-(x+1)/(x-2)=(x-2m)/[(x-2)(x+3)]同时乘以(x-2)(x+3),得x(x-2)-(x
如果问题是:1/2(x的平方+y的平方)-xy=2
2/(x-y)-(x+y)/(x-y)^2=[(2x-2y)-(x+y)]/(x-y)^2=(x-3y)/(x-y)^2
(一)25(x^2)/4-9=0,解移项得25(x^2)/4=9,两边同乘4/25得,x^2=36/25,开方得x=6/5.(二)x^2+8x+15=0,因式分解得(x+3)(x+5)=0,得x=-3
x^4-4x^2-x+2=x^2(x^2-4)-(x-2)=x^2(x+2)(x-2)-(x-2)=(x-2)(x^3+2x^2-1)=(x-2)(x+1)(x^2+x-1)
x/(2-x)-(2+x).(2-x)/((x+2)(x+2))=x/(2-x)-(2-x)/(x+2)=(6x-4)/(4-xx
x/(x-3)-(x+6)/(x^2-3x)+1/x=[x^2-(x+6)+(x-3)]/x(x-3)=(x^2-9)/x(x-3)=(x+3)/x.
x^2+2x-m+1=0没有实数根,所以:b^2-4ac=2^2+4(m-1)<0,即m<0x^2+mx+2x+2m+1=0x^2+(m+2)x+(2m+1)=0b^2-4ac=(m+2)^2-4(2