用适当的方法解方程(1)x2+3x-4=0 &nb
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用适当的方法解方程
(1)x2+3x-4=0
(2)4(x-3)2-x(x-3)=0
(3)3x2+12x=5(x+1)-8.
(1)x2+3x-4=0
(2)4(x-3)2-x(x-3)=0
(3)3x2+12x=5(x+1)-8.
![用适当的方法解方程(1)x2+3x-4=0 &nb](/uploads/image/z/8292202-34-2.jpg?t=%E7%94%A8%E9%80%82%E5%BD%93%E7%9A%84%E6%96%B9%E6%B3%95%E8%A7%A3%E6%96%B9%E7%A8%8B%EF%BC%881%EF%BC%89x2%2B3x-4%3D0%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nbsp%3B%26nb)
(1)x2+3x-4=0,
(x-1)(x+4)=0,
x1=1或x2=-4;
(2)4(x-3)2-x(x-3)=0,
(x-3)(4x-12-x)=0,
x-3=0或4x-12-x=0,
x1=3或x2=4;
(3)3x2+12x=5(x+1)-8,
3x2+12x-5x-5+8=0,
3x2+7x+3=0,
x=
−b±
b2−4ac
2a=
−7±
13
2×3,
x1=
−7+
13
6,x2=
−7−
13
6.
(x-1)(x+4)=0,
x1=1或x2=-4;
(2)4(x-3)2-x(x-3)=0,
(x-3)(4x-12-x)=0,
x-3=0或4x-12-x=0,
x1=3或x2=4;
(3)3x2+12x=5(x+1)-8,
3x2+12x-5x-5+8=0,
3x2+7x+3=0,
x=
−b±
b2−4ac
2a=
−7±
13
2×3,
x1=
−7+
13
6,x2=
−7−
13
6.