二次根式、一元二次方程题1、当a=[根号3]-1/[根号3]+1 ,求代数式2a^2-3a-4的值.2、已知:x=1/[

2个回答

  • 1.

    a = (√3 - 1)/(√3 + 1) = 2 - √3

    a^2 = 7 - 4√3

    ∴2a^2 - 3a - 4 = (14 - 8√3) + (3√3 - 6) - 4 = 4 - 5√3

    2.

    x = 1/[√5 + √3] = √5 - √3 ,y = 1/[√5 - √3] = √5 + √3 ,

    ∴(x + 2)(y + 2) = xy + 2(x + y) + 4 = 2 + 4√5 + 4 = 6 + 4√5

    3.

    [x - (2 + √2)][x - (2 - √2)]

    4.

    “2x^2-3x+m+1可以在实数范围内分解因式”等价于:方程 2x^2 - 3x + m + 1 = 0 有实数根 ,∴△》0 ,∴9 》8(m + 1) ,∴m《 1/8

    5.

    9x^2+6x+1-3y^2

    = (3x + 1)^2 - (√3y)^2

    = (3x + 1 + √3y)(3x + 1 - √3y)

    6.

    利用求根公式:△ = (-2√5)^2 - 4·2·1 = 12 ,√△ = 2√3 ,

    ∴x1 = (-b + √△)/2a = (√5 + √3)/2

    x2 = (-b - √△)/2a = (√5 - √3)/2