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九年級數學題,答案有,要過程

1、x?-10x+21=0

(x-3)(x-7)=0

x1=3,x2=7

2、原方程等於:

X^2 - 2*(1/2)*X + (1/2)^2 - (1/2)^2 - 1 = 0

=> (X - 1/2)^2 - 5/4 = 0

=> (X - 1/2)^2 = 5/4

=> X - 1/2 = ±√(5/4)

=> X1 = √(5/4) + 1/2

X2 = -√(5/4) + 1/2

=> X1 = (1+√5)/2

=> X2 = (1-√5)/2

3、壹元二次方程式ax^2+b+c=0

檢驗有沒有解:

b^2-4ac>0本題中有兩個不相等的實數解

3x^2+6x-4=0

x^2+2x-4/3=0

x^2+2x+1-1-4/3=0

(x+1)^2-7/3=0

(x+1)^2=7/3

(x+1)=±√21/3

x=±√21/3-1

4、3x(x+1)=3x+3

3x^2+3x=3x+3

3x^2+3x-3x-3=0

3(x^2-1)=0

x^2-1=0

x^2=1

x=±1

5、 4x^2-4x+1=x^2+6x+9

3x^2-10x-8=0

x^2-10x/3-8/3=0

(x-10/6)^2-25/9-8/3=0(x-10/6)^2=49/9

x-5/3=±7/3

x=±7/3+5/3

x=4或 -2/3

6、7x?-√6x-5=0

用公式解:

x=[-b±√(b?-4ac)]/2a

x=(√6±√146)/14