用因式分解解下列方程2(t-1)눀+t=1⑵(2x+1)눀-x눀=0⑶(x-5)눀=2(5-x)⑷

2024-11-25 05:24:43
推荐回答(1个)
回答1:

⑴2(t-1)²+t=1
解:2(t-1)²+t-1=0
(t-1)[2(t-1)+1]=0
t-1=0 2t-1=0
t1=1 t2=1/2
⑵(2x+1)²-x²=0
解:[(2x+1)+x][(2x+1)-x]=0
3x+1=0 x+1=0
x1=-1/3 x2=-1
⑶(x-5)²誉弯=2(5-x)
解:(x-5)²-2(5-x)=0
(x-5)²返神+2(x-5)=0
(x-5)(x-5+2)=0
x-5=0 x-3=0
x1=5 x2=3
⑷x²-9x+20=0
解:(x-4)(x-5)=0
x-4=0 x-5=0
x1=4 x2=5
⑸9(x-1)²-4=0
解:[3(x-1)+2][3(x-1)-2]=0
3x-1=0 3x-5=0
x1=1/3 x2=5/庆世闷3