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

Toán Lớp 6: Tìm x, y ,z : Xy = 3/5 ; yz = 4/5 ; xz = 3/4

Toán Lớp 6: Tìm x, y ,z :
Xy = 3/5 ; yz = 4/5 ; xz = 3/4

Comments ( 2 )

  1. Ta có : xy.yz.xz=3/5 . 4/5 . 3/4
    to (xyz)^2=9/25
    to \(\left[ \begin{array}{l}xyz=\dfrac{3}{5}\\xyz=-\dfrac{3}{5}\end{array} \right.\) 
    Với xyz=3/5
    to
    z=3/5 : xy=3/5:3/5=1
    x=4/5 : yz=3/5:4/5=3/4
    y=4/5 : xz=3/5:3/4=4/5
    Với xyz=-3/5
    to
    z
    z=-3/5 : xy=-3/5:3/5=-1
    x=-4/5 : yz=-3/5:4/5=-3/4
    y=-4/5 : xz=-3/5:3/4=-4/5
    Vậy (x;y;z)=(3/4;4/5;1);(-3/4;-4/5;-1)

  2. Giải đáp:
    $\\$
    xy = 3/5 (1)
    yz = 4/5 (2)
    xz = 3/4 (3)
    Lấy (1) . (2) . (3) ta được :
    -> xy . yz  . xz = 3/5 . 4/5 . 3/4
    -> (x.x) (y.y) (z.z) = 9/25
    -> x^2y^2z^2=9/25
    -> (xyz)^2=9/25
    -> \(\left[ \begin{array}{l}(xyz)^2=(\dfrac{3}{5})^2\\(xyz)^2=(\dfrac{-3}{5})^2\end{array} \right.\)
    -> \(\left[ \begin{array}{l}xyz=\dfrac{3}{5}\\xyz=\dfrac{-3}{5}\end{array} \right.\) 
    $\bullet$ Với xyz = 3/5 (4)
    Lấy (4) : (1) ta được :
    -> (xyz) : (xy) = 3/5 : 3/5
    -> z = 1
    Lấy (4) : (2) ta được :
    -> (xyz) : (yz) = 3/5 : 4/5
    -> x = 3/4
    Lấy (4) : (3) ta được :
    -> (xyz) : (xz) = 3/5 : 3/4
    -> y = 4/5
    $\bullet$ Với xyz=(-3)/5 (5)
    Lấy (5):(1) ta được :
    -> (xyz) : (xy) = (-3)/5 : 3/5
    -> z = -1
    Lấy (5) : (2) ta được :
    -> (xyz) : (yz) = (-3)/5 : 4/5
    -> x = (-3)/4
    Lấy (5) : (3) ta được :
    -> (xyz) : (xz) = (-3)/5 : 3/4
    -> y = (-4)/5
    Vậy (x;y;z) = ( 3/4; 4/5 ; 1), ( (-3)/4 ; (-4)/5; -1)
     

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

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