Question

用因式分解法解下列方程

1.（x-4)²+19（x-4)-20=0
2.(2x-5)²-9（x-2)²=0
3.2（t-1)²+t=0
4.3(1-x)²-5(x-1)-2=0
5.[x/(x²-x)]+2=2x/(x+1)
要详细的过程。谢谢了~！

Answer 1

1. ((x-4)+20)((x-4)-1)=0得x1=-16,x2=5
2.(2x-5-3x+6)(2x-5+3x-6)=0得x1=1,x2=11/5
3.2(t-1)²+(t-1)+1=0 △ =1²-4x2x1=-7无解
4.(3(x-1)+1)((x-1)-2)=0 x1=2/3,x2=3
5. 将2移到等式右边然后通分则 x/(x²-x）=2/（x+1） 在x不等于0时化简等式2x²-2x=x²+x 即x²-3x=0 得x1=0(舍)，x2=1/3

Answer 2

1.(x-4-1)(x-4+20)=0,x=5 or -16
2.4x²-20x+25-9x²+36x-36=0，5x²-16x+11=0，x=11/5 or 1
3.2t²-4t+2+t=0,2t²-3t+2=0,t=-1/2 or 2
4.[3(1-x)-1][(1-x)+2]=0,x=2/3 or 3
5.左右各乘以(x-1)(x+1)得：(x+1)+2(x²-1)=2x(x-1),x=2

Answer 3

1.（X-4+20)(X-4-1)=0 (X+15)(X-5)=0 X1=-15 X2=5
2. 4X²-20X+25-9X²+36X-36=0 -5X²+16X-11=0 (-5X-11)(X+1)=0
X1=-11/5 X=-1
3.2t²-4t+2+t=0 2t²-3t+2=0 用求根公式
4.3-6X+3X²-5X+5-2=0 3X²-11X+6=0 (X-3)(3X-2)=0 X1=3 X2=2/3
5.1/（X-1)+2=2X/(X+1)
两边乘于X²-1 X+1+2X²-2=2X(X-1) 3X-1=0 X=1/3