用因式分解解方程(有过程):(1)9y^2=(y-z)^2 (2)(4x-1)^2=(1-5x)^2 (3)x^3-4x^2+4x=0

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 19:22:21
用因式分解解方程(有过程):(1)9y^2=(y-z)^2 (2)(4x-1)^2=(1-5x)^2 (3)x^3-4x^2+4x=0
xݓj0_Eٱ]H{6 emʺinkaڔuK;ŒݫB?qdf1HGt:`Oxzŷ|vn{Gջ#vڈqbE#J KUj~Τ-8`F[዗:ӯ:|OÏm c;J(H*Hr$2$4cW/ S%4,E 0)6%0mdm` hkA1ZXuTu6=+~a,n?ý}[s U#~QKsӹ>o8lM.Ee(g鞸9WdrFln%D06?.?c9l$c~pfKj S6-v5m1[砚D!x Uvz-tC=AbD:=h8ʯjPH k*\Swq7t0{w&V2ow%yfhڗ6bbYQT

用因式分解解方程(有过程):(1)9y^2=(y-z)^2 (2)(4x-1)^2=(1-5x)^2 (3)x^3-4x^2+4x=0
用因式分解解方程(有过程):(1)9y^2=(y-z)^2 (2)(4x-1)^2=(1-5x)^2 (3)x^3-4x^2+4x=0

用因式分解解方程(有过程):(1)9y^2=(y-z)^2 (2)(4x-1)^2=(1-5x)^2 (3)x^3-4x^2+4x=0
9y^2=(y-z)^2
9y^2-(y-z)^2=0
(3y+y-z)(3y-y+z)=0
(4y-z)(2y+z)=0
4y-z=0.2y+z=0
y=z/4,y=-z/2
(2)(4x-1)^2=(1-5x)^2
(4x-1)^2-(1-5x)^2=0
(4x-1+1-5x)(4x-1-1+5x)=0
(-x)(9x-2)=0
-x=0,9x-2=0
x=0,x=2/9
(3)x^3-4x^2+4x=0
x(x^2-4x+4)=0
x(x-2)^2=0
x=0,x-2=0
x=0,x=2
你能明白,赞同

yun

第(1)条方程有2个未知数,没有唯一解
第(2)条方程不用因式分解,直接用4x-1=±(1-5x)就可以了
第(3)条方程可分解为:x(x-2)^2=0,解为0或2

1. 9y^2-(y-z)^2=0
[3y+(y-z)][3y-(y-z)] =0
(4y-z)(2y+z)=0
4y-z=0 y1=1/4z
2y+z=0 y2=-1/2z
2. (4x-1)^2-(1-5x)^2=0
(4x-1+ 1-5x)[4x-1-(1-5x)]=0
...

全部展开

1. 9y^2-(y-z)^2=0
[3y+(y-z)][3y-(y-z)] =0
(4y-z)(2y+z)=0
4y-z=0 y1=1/4z
2y+z=0 y2=-1/2z
2. (4x-1)^2-(1-5x)^2=0
(4x-1+ 1-5x)[4x-1-(1-5x)]=0
-x*(9x+2)=0
x1=0
x2=-2/9
3. x^3-4x^2+4x=0
x(x^2-4x+4)=0
x(x-2)^2=0
x1=0 x2=x3=2

收起