Question

解方程~~!

解方程 , 我还没学过~~

1.(3y^2-y)^2=3(3y^2-y)-2
2.(t^2+t-1)(t^2+t+2)=4
3.x^2-4ax+4a^2-b^2=0
4.x^2-m(3x-2m+n)=n^2
5.x^2-|x-1|-1=0

Answer 1

1.(3y^2-y)^2=3(3y^2-y)-2
令a=3y^2-y
a^2-3a+2=0
(a-1)(a-2)=0
a=1,a=2

3y^2-y=a=1
3y^2-y-1=0
y=(1±√13)/6

3y^2-y=a21
3y^2-y-2=0
(y-1)(3y+2)=0
y=1,y=-2/3

y=(1+√13)/6,y=(1-√13)/6,y=1,y=-2/3

2.(t^2+t-1)(t^2+t+2)=4
令x=t^2+t
(x-1)(x+2)=4
x^2+x-2=4
x^2+x-6=0
(x-2)(x+3)=0
x=2,x=-3

t^2+t=2
(t-1)(t+2)=0
t=1,t=-2

t^2+t=-3
t^2+t+3=0
判别式小于0，无解

所以t=1,t=-2

3.x^2-4ax+4a^2-b^2=0
(x-2a)^2=b^2
x-2a=±b
x=2a+b,x=2a-b

4.x^2-m(3x-2m+n)=n^2
x^2-3mx+2m^2-mn-n^2=0
(x-m)(x-2m)-mn-n^2=0
(x-m)(x-2m)+nx-nx-mn-n^2=0
(x-m)(x-2m)-n(x-m)+nx+2mn-n^2=0
(x-m)(x-2m-n)+n(x-2m-n)=0
(x-m+n)(x-2m-n)=0
x=m-n,x=2m+n

5.x^2-|x-1|-1=0
若x>=1,|x-1|=x-1
x^2-x=0
x(x-1)=0
x>=1,所以x=1

若x<0,|x-1|=1-x
x^2-1+x-1=x^2+x-2=0
(x+2)(x-1)=0
x<0，所以x=-2
所以x=1,x=-2

Answer 2

Answer 3

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