方程2x-1-x²-1=-2分之1的实根
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/01 13:16:44
(3x-2)(5x+1)=0,✘1=-1/5,✘2=2/3再问:给个过程再答:这不是过程吗?再答:后面的提2出来,再答:在移项再问:咋变的再答:2(5x+1)再问:好了我会
(2x+1)²=3(2x+1)(2x+1)²-3(2x+1)=0(2x+1)[(2x+1)-3]=0(2x+1)(2x-2)=02x+1=0或2x-2=0x=-1/2或x=1
解题思路:此题主要考查了因式分解法解一元二次方程,正确分解因式是解题关键.解题过程:解:(2x-1)2=3x(2x-1)(2x-1)2-3x(2x-1)=0(2x-1)[(2x-1)-3x]=0(2x
再答:看看可以不再答:看看可以不再问:嗯,谢谢再答:不会了来问我啊,当年哦高考考了138
移项4x(2x+1)-3(2x+1)=0提取公因式(2x+1)(4x-3)=0
1/2x²+x+1/2=0x²+2x+1=0(x+1)²=0解得:x=-1
原方程就是X^2-5x-3=0X^2-5x+25/4-37/4=0(x-5/2)^2-(√37/2)^2=0(x-5/2+√37/2)(x-5/2-√37/2)=0x=5/2-√37/2或者x=5/2
2x(5x-1)=3(5x-1)2x(5x-1)-3(5x-1)=0(5x-1)(2x-3)=05x-1=0或2x-3=0得x=1/5或x=3/2如还不明白,请继续追问.手机提问的朋友在客户端右上角评
x的解为-2/3或-4移项(x-1)²=(2x+3)²(x-1)²-(2x+3)²=0得(x-1+2x+3)(x-1-2x-3)=0因此x-1+2x+3=0,3
(2x+1)²=2(2x+1)(2x+1)²-2(2x+1)=0(2x+1)(2x+1-2)=0(2x+1)(2x-1)=0x1=-1/2,x2=1/2再问:(2x+1)²
3x(x-1)=2(x-1)3x²-3x=2x-23x²-5x+2=0(3x-2)(x-1)=03x-2=0,x-1=0所以x=2/3或x=1再问:3x²-5x+2=0(
3x(x-1)+2(x-1)=0,(x-1)(3x+2)=0,x-1=0或3x+2=0,所以x1=1,x2=-23.
(4x-3)(3x-1)=212x^2-4x-9x+3-2=012x^2-13x+1=0(12x-1)(x-1)=0x1=1/12x2=1(3x-2)=(2-3x)(x+1)(3x-2)+(3x-2)
移项4x²-1=0(2x+1)(2x-1)=0x=-1/2,x=1/2
(3x+1)(x-1)=(2x-5)(x-1)(3x+1)(x-1)-(2x-5)(x-1)=0(x-1)(3x+1-2x+5)=0(x-1)(x+6)=0x=1,x=-6
2x(2x+5)=(x-1)(2x+5)4x²+10x=2x²+3x-52x²+7x+5=0(2x+5)(x+1)=0x=-5/2或x=-1
(5x-3)^2=(2x+1)^25x-3=2x+1或5x-3=-2x-13x=4或7x=2x=4/3或x=2/73(x+3)^2=x^2+x3x²+18x+27=x²+x2x
(x-1)(2x+3)+x(1-x)=0(x-1)(2x+3)-x(x-1)=0(x-1)(2x+3-x)=0(x-1)(x+3)=0x-1=0x=1x+3=0x=-3
2(x+1)^2+3(x+1)=0(x+1)[2(x+1)+3]=0;(x+1)(2x+5)=0;x=-1或x=-5/2;很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追