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

Toán Lớp 7: Tìm x,y,z biết: $\frac{x}{y+z+1}$ =$\frac{y}{x+z+2}$ =$\frac{z}{x+y-3}$ =x+y+z

Toán Lớp 7: Tìm x,y,z biết:
$\frac{x}{y+z+1}$ =$\frac{y}{x+z+2}$ =$\frac{z}{x+y-3}$ =x+y+z

Comments ( 1 )

  1. ( * ) Nếu x , y , z $\neq$ 0 
    Áp dụng tính chất dãy tỉ số bằng nhau ta có :
    $\frac{x}{y+z+1}$ = $\frac{y}{x+z+2}$ = $\frac{z}{x+y-3}$ = $\frac{x+y+z}{y+z-1+x+z+2+x+y-3}$ = $\frac{x+y+z}{x+y+z+x+y+z}$ = $\frac{x+y+z}{2( x+y + z )}$ = $\frac{1}{2}$
    ⇒ x + y + z = $\frac{1}{2}$ 
    x + y = $\frac{1}{2}$ – z
    x + z = $\frac{1}{2}$ – y
    y + z = $\frac{1}{2}$ – x
    ( 1 ) Thay y + x + 1 = $\frac{1}{2}$ – x + 1
    ⇒ $\frac{x}{\frac{1}{2}-x+1}$ = $\frac{1}{2}$
    2x = $\frac{1}{2}$ – x + 1
    ⇒ 3x = $\frac{1}{2}$ + 1 = $\frac{3}{2}$ 
    x = $\frac{3}{2}$ : 3 = $\frac{1}{2}$ 
    ( 2 ) Thay x + z + 2 = $\frac{1}{2}$ – y + 2
    ⇒ $\frac{y}{\frac{1}{2} – y + 2 }$ = $\frac{1}{2}$ 
    3y = $\frac{1}{2}$ + 2 = $\frac{5}{2}$ 
    y = $\frac{5}{2}$ : 3 = $\frac{5}{6}$ 
    ( 3 ) Thay x + y – 3 = $\frac{1}{2}$ – z + 3
    ⇒ $\frac{z}{\frac{1}{2} – z – 3 }$ = $\frac{1}{2}$ 
    3z = $\frac{1}{2}$ – 3 = $\frac{-5}{2}$ 
    z = $\frac{-5}{2}$ : 3 = $\frac{-5}{6}$ 
    ⇔ x = $\frac{1}{2}$ ; y = $\frac{5}{6}$ và z = $\frac{-5}{6}$ 
    TH2 : x + y + z = 0
    ⇒ $\frac{x}{y+z+1}$ = $\frac{y}{x+z+2}$ = $\frac{z}{x+y-3}$ = 0
    ⇔ x = y = z = 0 

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