一元二次方程因式分解法(1)——x^2-2x-99=0 (2)——2y(y-4)=4-y (3)——(x-5)^2+2x(x-5)=0 (4)4(x-√2）^2=3(√2-x)
问题描述：
一元二次方程因式分解法
(1)——x^2-2x-99=0 (2)——2y(y-4)=4-y (3)——(x-5)^2+2x(x-5)=0 (4)4(x-√2）^2=3(√2-x)
答
(1) x^2-2x-99=0
x^2-2x+1-100=0
(x-1)^2-100=0
(x-1)^2=100
x1=11 x2=-9
(2) 2y(y-4)=4-y
2y^2-7y-4=0
(x-4)(2x+1)=0
x1=4 x2=-0.5
(3) (x-5)^2+2x(x-5)=0
(x-5)(x-5+2x)=0
(x-5)(3x-5)=0
x1=5 x2=5/3
(4) 4(x-√2）^2=3(√2-x)
4(x-√2）^2+3(x-√2)=0
(x-√2)(4x-4√2+3)=0
x1=√2 x2=√2-3/4
不放心的话您可以检查一下哦O(∩_∩)O
答
(1) x^2-2x-99=0 x^2-2x+1-100=0(x-1)^2-100=0(x-1)^2=100x1=11 x2=-9(2) 2y(y-4)=4-y2y^2-7y-4=0(x-4)(2x+1)=0x1=4 x2=-0.5(3) (x-5)^2+2x(x-5)=0(x-5)(x-5+2x)=0(x-5)(3x-5)=0x1=5 x2=5/3(4) 4(x-√2）^2=3(√2-x)...