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

Toán Lớp 7: cho x/a = y/b = z/c C/M bz-cy/a = cx-az/b = ay-bx/c

Toán Lớp 7: cho x/a = y/b = z/c C/M bz-cy/a = cx-az/b = ay-bx/c

Comments ( 2 )

  1. $\\$
    Giả sử : (bz – cy)/a = (cx – az)/b = (ay – bx)/c, ta chứng minh : x/a = y/b=  z/c
    Có : (bz – cy)/a = (cx – az)/b = (ay – bx)/c 
    -> (x (bz – cy) )/(ax) = (y (cx – az) )/(by) = (z (ay – bx) )/(cz)
    -> (bxz – cyz)/(ax) = (cyz – ayz)/(by) = (ayz – bxz)/(cz)
    Áp dụng tính chất dãy tỉ số bằng nhau có :
    (bxz – cyz)/(ax) = (cyz – ayz)/(by) = (ayz – bxz)/(cz) = (bxz – cyz + cyz – ayz + ayz – bxz)/(ax + by + cz) = ( (bxz – bxz) + (-cyz + cyz) + (-ayz + ayz) )/(ax + by + cz) = 0/(ax + by + cz)=0
    -> $\begin{cases} \dfrac{bz-cy}{a}=0\\\dfrac{cx-az}{b}=0\\\dfrac{ay – bx}{c}=0\end{cases}$ -> $\begin{cases} bz – cy=0\\cx-az=0\\ay – bx=0 \end{cases}$ -> $\begin{cases} bz=cy\\cx=az\\ay=bx\end{cases}$
    Có : bz = cy
    -> b/c = y/z
    -> b/y = c/z
    -> y/b = z/c (*)
    Có : cx = az
    -> c/a =z/x
    -> c/z = a/x
    -> z/c = x/a (**)
    Từ (*), (**)
    -> x/a = y/b = z/c 
    Từ đó ta khẳng định :
    Nếu x/a=y/b=z/c thì (bz-cy)/a = (cx-az)/b = (ay-bx)/c

  2. Đặt x/a = y/b = z/c = k
    Suy ra :
    {(x= ak ),(y = bk ),(z= ck):}
    Khi đó ta có :
    (bz – cy)/a = (b . ck – c . bk)/a = 0/a = 0 (1)
    (cx -az)/b = (c . ak – a . ck)/b = 0/b = 0 (2)
    (ay – bx)/c = (a . bk – b . ak)/c = 0/c = 0 (2)
    Từ (1) ; (2) ; (3) ta có :
    (bz- cy)/a = (cx – az)/b = (ay -bx)/c
    Vậy với x/a = y/b = z/c thì (bz- cy)/a = (cx – az)/b = (ay -bx)/c

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