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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 10: giải pt : x mũ 4 – 6x mũ 3 + 12x mũ 2 – 9x + 2 = 0

Tháng Tám 26, 2022 Toán học 1 Comment 0 views

Toán Lớp 10: giải pt : x mũ 4 – 6x mũ 3 + 12x mũ 2 – 9x + 2 = 0

Comments ( 1 )

  1. aivilanlan
    aivilanlan
    26 Tháng Tám, 2022 at 8:45 sáng
    Reply
    x^4-6x^3+12x^2-9x+2=0
    <=>x^4-6x^3+12x^2-8x-x+2=0
    <=>x(x^3-6x^2+12x-8)-(x-2)=0
    <=>x[x^3-3.x^2 .2+3.x.2^2 -2^3]-(x-2)=0
    <=>x(x-2)^3-(x-2)=0
    <=>(x-2)[x(x-2)^2-1]=0
    <=>[(x-2=0),(x(x-2)^2-1=0):}
    <=>[(x=2),(x(x^2-4x+4)-1=0):}
    <=>[(x=2),(x^3-4x^2+4x-1=0):}
    <=>[(x=2),(x^3-1-4x^2+4x=0):}
    <=>[(x=2),((x-1)(x^2+x+1)-4x(x-1)=0):}
    <=>[(x=2),((x-1)(x^2+x+1-4x)=0):}
    <=>[(x=2),(x=1),(x^2-3x+1=0):}
    <=>[(x=2),(x=1),(x^2-2.x. 3/2+9/4-5/4=0):}
    <=>[(x=2),(x=1),((x-3/2)^2-5/4=0):}
    <=>$\left[\begin{matrix} x=2\\x=1\\(x-\frac{3}{2}-\frac{\sqrt{5}}{2})(x-\frac{3}{2}+\frac{\sqrt{5}}{2})=0\end{matrix}\right.$
    <=>[(x=2),(x=1),((x-\frac{\sqrt{5}+3}{2}),(x-\frac{3-\sqrt{5}}{2}=0)):}
    <=>[(x=2),(x=1),(x=\frac{+-\sqrt{5}+3}{2}):}
    Vậy S={2;1;\frac{+-\sqrt{5}+3}{2}}

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222-9+11+12:2*14+14 = ? ( )

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