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

Toán Lớp 8: tìm x để x(x+2)^2+x(x-5)-5(x+2)^2=0

Tháng Sáu 7, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: tìm x để x(x+2)^2+x(x-5)-5(x+2)^2=0

Comments ( 1 )

  1. lanngocha
    lanngocha
    7 Tháng Sáu, 2022 at 5:29 sáng
    Reply
    x(x + 2)^2 + x(x – 5) – 5(x + 2)^2 = 0
    ⇔(x-5)(x+2)^2+x(x-5)=0
    ⇔(x-5)(x^2+4x+4)+x(x-5)=0
    ⇔(x-5)(x^2+4x+4+x-5)=0
    ⇔(x-5)(x^2+5x-1)=0
    TH1: x-5=0⇔x=5
    TH2:
    x^2+5x-1=0
    ⇔x^2+2. 5/2.x+25/4-25/4+1=0
    ⇔(x-5/2)^2-21/4=0
    ⇔(x-5/2)^2=21/4
    ⇔x-5/2=±(sqrt{21})/2
    ⇔x=±(sqrt{21})/2+5/2
    Vậy x∈{5; ±(sqrt{21})/2+5/2}.
     

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

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