Question

用配方法解4x^-8x+3=0急！！

Answer 1

解：4x^2-8x+3=0 （先两边同时除以二次项系数）
x^2-2x+3/4=0
x^2-2x+1-1+3/4=0
(x-1)^2=1/4
x-1=±1/2
x-1=1/2或x-1=-1/2
解得x1=3/2，x2=1/2

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Answer 2

4x²-8x+3=0
4x²-8x+4=1
x²-2x+1=1/4
(x-1)²=1/4
x-1=1/2或x-1=-1/2
x=3/2或x=1/2
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Answer 3

解：
4x²-8x+3=0
4(x²-2x+3/4)=0
4(x²-2x+1-1+3/4)=0

4[(x-1)²-1/4]=0
(x-1)²=1/4

x-1=±1/2
x=1±(1/2)
因此：原函数的根为：3/2和1/2

Answer 4

4x^2-8x+3=0
4（x^2-2x）+3=0

4（x^2-2x+1-1）+3=0

4（x^2-2x+1）-4+3=0

4（x^2-2x+1）=1

（x^2-2x+1）=1/4

（x-1）^2=1/4
x-1=1/4 x-1=-1/4

x=3/2或x=1/2

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Answer 5

4x^2-8x+3=0
4(x^2-2x+1)-4+3=0
4(x-1)^2=1
(x-1)^2=1/4
x-1=1/2 x=3/2
x-1=-1/2 x=1/2