解一元二次方程①x2-x-12=0 &
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解一元二次方程
①x2-x-12=0
②(x+1)(x-2)=x+1.
①x2-x-12=0
②(x+1)(x-2)=x+1.
![解一元二次方程①x2-x-12=0 &](/uploads/image/z/19545616-64-6.jpg?t=%E8%A7%A3%E4%B8%80%E5%85%83%E4%BA%8C%E6%AC%A1%E6%96%B9%E7%A8%8B%E2%91%A0x2-x-12%3D0%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26)
①x2-x-12=0,
分解因式得:(x+3)(x-4)=0,
可得x+3=0或x-4=0,
解得:x1=-3,x2=4.
②(x+1)(x-2)-(x+1)=0,
∴(x+1)(x-2-1)=0,即(x+1)(x-3)=0,
∴x+1=0,或x-3=0,
∴x1=-1,x2=3.
分解因式得:(x+3)(x-4)=0,
可得x+3=0或x-4=0,
解得:x1=-3,x2=4.
②(x+1)(x-2)-(x+1)=0,
∴(x+1)(x-2-1)=0,即(x+1)(x-3)=0,
∴x+1=0,或x-3=0,
∴x1=-1,x2=3.