解方程式

√(5x^2-6x+1)+√(5x^2+9x-2)=5x-1
==>√[(5x-1)(x-1)]+√[(5x-1)(x+2)]=5x-1
令5x-1=y
√[y(x-1)]+√[y(x+2)]=y
==>y(x-1)+2√[y^2(x-1)(x+2)]+√(x-2)=y^2
yx-y+2y√(x-1)(x+2)]+√(x-2=y^2)
==>x-1+2√[(x-1)(x+2)]+x+2=y
y已5x-1帶回去
2x+1+2√[(x-1)(x+2)]=5x-1
2√[(x-1)(x+2)]=3x-2
==>4(x-1)(x+2)=(3x-2)^2
4(x^2+2x-x-2)=9x^2-12x+4
4x^2+4x-8=9x^2-12x+4
5x^2-16x+12=0
(5x-6)(x-2)=0
x=6/5 or 2

不知道對不對^^

√(5x^2-6x+1)+ √(5x^2+9x-2)=5x-1
√[(5x-1)(x-1)]+√[(5x-1)(x+2)]=5x-1
(√(5x-1))(√(x-1)+√(x+2))=5x-1
√(x-1)+√(x+2)=√(5x-1)
2x+1+2√((x-1)(x+2))=5x-1
2√((x-1)(x+2))=3x-2
4x^2+4x-8=9x^2-12x+4
5x^2-16x+12=0
x=(16±4)/10
x=2 or 6/5

代回原方程式(這是解根式方程很重要的一步!)
皆符合

√(5x^2-6x+1)+ √(5x^2+9x-2)=5x-1
√[(5x-1)(x-1)]+√[(5x-1)(x+2)]=5x-1
√(x-1)+√(x+2)=√(5x-1)
2x+1+2√((x-1)(x+2))=5x-1
2√((x-1)(x+2))=3x-2
5x^2-16x+12=0
x=2 or 6/5

√(5x^2-6x+1)+√(5x^2+9x-2)=5x-1
√[(5x-1)(x-1)]+√[(5x-1)(x+2)]=5x-1
令5x-1=A
√[A(x-1)]+√[A(x+2)]=A
A(x-1+X+2)+2√[A^2(x-1)(x+2)]=A^2
2Ax+A+2A√(x-1)(x+2)]=A^2
2x+1+2√(x^2+x-2)=A
帶入A=5x-1
5x^2-16x+12=0
x=2或6/5

√(5x-1)(x-1)+√(5x-1)(x+2)=5x-1
令y=5x-1
√y(x-1)+√y(x+2)=y
兩邊平方
y(x-1)+2√y^2(x-1)(x+2)+y(x+2)=y^2
y(2x+1)+2√y^2(x-1)(x+2)=y^2
2√y^2(x-1)(x+2)=y(y-2x-1)
代入y=5x-1
2√y^2(x-1)(x+2)=y(3x-2)
兩邊平方
4y^2(x-1)(x+2)=y^2(9x^2-12x+4)
y^2[4(x-1)(x+2)-(9x^2-12x+4)]=0
代入y=5x-1
(5x-1)^2(5x^2-16x+12)=0
(5x-1)^2(x-2)(5x-6)=0
x=1/5, 2, 5/6

有三解1/5, 2, 5/6

等式兩邊不可任意消去某ㄧ公因式
否則會發生減根的情形
例如:
(5x-1)f(x)=(5x-1)g(x)
此方程式本來有ㄧ根為x=1/5,但若先同時除以5x-1,則解出來的根將缺少x=1/5這個根
所以上述例子應先將5x-1提出
即(5x-1)[f(x)-g(x)]=0
則5x-1=0, f(x)-g(x)=0