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

Toán Lớp 8: Tìm x,y,z biết : 9x^2 + y^2 +2z^2 -18x + 4z -6y +20 = 0

Tháng Sáu 20, 2022 Toán học 2 Comments 0 views

Toán Lớp 8: Tìm x,y,z biết : 9x^2 + y^2 +2z^2 -18x + 4z -6y +20 = 0

Comments ( 2 )

  1. tongtung
    tongtung
    20 Tháng Sáu, 2022 at 1:09 chiều
    Reply
    9x^2 + y^2 + 2z^2 – 18x + 4z – 6y + 20 = 0
    (9x^2 – 18x + 9) + (y^2 – 6y + 9) + (2z^2 + 4z + 2) = 0
    (3x – 3)^2 + (y – 3)^2 + 2(z + 1)^2 = 0
    => [(3x – 3 = 0),(y – 3 = 0),(2(z + 1) = 0):}
    => [(x = 1),(y = 3),(z = -1):}
    Gửi bạn
     

  2. dinhcongson
    dinhcongson
    20 Tháng Sáu, 2022 at 1:08 chiều
    Reply
    Bạn tham khảo nhé.
    9x^2+y^2+2z^2-18x+4z-6y+20=0
    <=>9x^2-18x+9+y^2-6y+9+2z^2+4z+2=0
    <=>[(3x)^2-2.3x.3+3^2]+[y^2-2.y.3+3^2]+2[z^2+2.z.1+1^2]=0
    <=>(3x-3)^2+(y-3)^2+2(z+1)^2=0
    Ta có: {((3x-3)^2>=0),((y-3)^2>=0),(2(z+1)^2>=0):}AAx,y,z
    \text{Để}:(3x-3)^2+(y-3)^2+2(z+1)^2=0
    =>{((3x-3)^2=0),((y-3)^2=0),(2(z+1)^2=0):}
    <=>{(3x-3=0),(y-3=0),((z+1)^2=0):}
    <=>{(3x=3),(y=3),(z+1=0):}
    <=>{(x=1),(y=3),(z=-1):}
    Vậy (x;y;z)=(1;3;-1)
     

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

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