x²+16x=0,用因式分解法解二元一次方程(1)x²+16x=0(2)5x平方-10x=-5(3)x(x-3)+x-3=0(4)2(x-3)²=9-x²用因式分解法解二元一次方程
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![x²+16x=0,用因式分解法解二元一次方程(1)x²+16x=0(2)5x平方-10x=-5(3)x(x-3)+x-3=0(4)2(x-3)²=9-x²用因式分解法解二元一次方程](/uploads/image/z/74563-43-3.jpg?t=x%26%23178%3B%2B16x%3D0%2C%E7%94%A8%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E6%B3%95%E8%A7%A3%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B%EF%BC%881%EF%BC%89x%26%23178%3B%2B16x%3D0%EF%BC%882%EF%BC%895x%E5%B9%B3%E6%96%B9-10x%3D-5%EF%BC%883%EF%BC%89x%EF%BC%88x-3%EF%BC%89%2Bx-3%3D0%EF%BC%884%EF%BC%892%EF%BC%88x-3%EF%BC%89%26%23178%3B%3D9-x%26%23178%3B%E7%94%A8%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E6%B3%95%E8%A7%A3%E4%BA%8C%E5%85%83%E4%B8%80%E6%AC%A1%E6%96%B9%E7%A8%8B)
x²+16x=0,用因式分解法解二元一次方程(1)x²+16x=0(2)5x平方-10x=-5(3)x(x-3)+x-3=0(4)2(x-3)²=9-x²用因式分解法解二元一次方程
x²+16x=0,用因式分解法解二元一次方程
(1)x²+16x=0
(2)5x平方-10x=-5
(3)x(x-3)+x-3=0
(4)2(x-3)²=9-x²
用因式分解法解二元一次方程
x²+16x=0,用因式分解法解二元一次方程(1)x²+16x=0(2)5x平方-10x=-5(3)x(x-3)+x-3=0(4)2(x-3)²=9-x²用因式分解法解二元一次方程
1)x²+16x=0
x(x+16)=0
x=0或x=-16
(2)5x平方-10x=-5
x方-2x+1=0
(x-1)方=0
x1=x2=1
(3)x(x-3)+x-3=0
(x-3)(x+1)=0
x1=3,x2=-1
(4)2(x-3)²=9-x²
2(x-3)方+(x+3)(x-3)=0
(x-3)[2(x-3)+x+3]=0
(x-3)(3x-3)=0
(x-3)(x-1)=0
x1=3或x2=1