2(t-1)的平方+t=1
2(t-1)²+t-1=0
(t-1)(2t-2+1)=0
(t-1)(2t-1)=0
t=1 t=1/2
4(x-3)的平方-25(x-2)的平方=0
(x-3+5x-10)(x-3-5x+10)=0
(6x-13)(-4x+7)=0
∴x=13/6 x=7/4
x的平方+2(a+1)乘x+(a的平方+2a+1)=0
x²+2(a+1)x+(a+1)²=0
(x+a+1)²=0
x=-a-1
2(t-1)的平方+t=1
2(t-1)²+t-1=0
(t-1)(2t-2+1)=0
(t-1)(2t-1)=0
t=1 t=1/2
4(x-3)的平方-25(x-2)的平方=0
(x-3+5x-10)(x-3-5x+10)=0
(6x-13)(-4x+7)=0
∴x=13/6 x=7/4
x的平方+2(a+1)乘x+(a的平方+2a+1)=0
x²+2(a+1)x+(a+1)²=0
(x+a+1)²=0
x=-a-1