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

Toán Lớp 8: Tìm x (2x^4-3x^3+x^2):(x^2)+4(x+1)^2=0

Tháng Chín 10, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Tìm x
(2x^4-3x^3+x^2):(x^2)+4(x+1)^2=0

Comments ( 1 )

  1. hongocha12
    hongocha12
    10 Tháng Chín, 2022 at 11:47 chiều
    Reply
    #tnvt
    (2x^4-3x^3+x^2):(-x^2)+4(x+1)^2=0
    <=>-2x^2+3x-1+4(x^2+2x+1)=0
    <=>-2x^2+3x-1+4x^2+8x+4=0
    <=>2x^2+11x+3=0
    <=>2x^2+2.2.x. 11/4+2. 121/16-2.97/16=0
    <=>2(x^2+2.x. 11/4+121/16-97/16)=0
    <=>(x+11/4)^2-(\frac{\sqrt{97}}{4})^2=0
    <=>(x+11/4-\frac{\sqrt{97}}{4)(x+11/4+\frac{\sqrt{97}}{4})=0
    <=>(x+\frac{11-\sqrt{97}}{4})(x+\frac{\sqrt{97}+11}{4})=0
    <=>$\left[\begin{matrix} x=\dfrac{-11+\sqrt{97}}{4}\\ x=\dfrac{-\sqrt{97}-11}{4}\end{matrix}\right.$
    Vậy S={\frac{-11+-\sqrt{97}}{4}}

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

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