初二数学题!急!在线等!1.已知A(Y+Z)=B(Z+X)=C(X+Y),A,B,C都不为零求证Y-Z/A(B-C)=Z

4个回答

  • 1、

    A(Y+Z)=B(Z+X)

    A=B(Z+X)/(Y+Z)

    C(X+Y)=B(Z+X)

    C=B(Z+X)/(X+Y)

    A-B=B[(Z+X)/(Y+Z)-1]

    =(X-Y)/(Y+Z)

    B-C=B[1-(Z+X)/(X+Y)

    =(Y-Z)/(X+Y)

    (Y-Z)/A(B-C)=(Y-Z)(X+Y)/A(Y-Z)=(X+Y)/A

    (X-Y)/C(A-B)=(X-Y)(Y+Z)/C(X-Y)=(Y+Z)/C

    由于A(Y+Z)=C(X+Y)

    因此,(X+Y)/A=(Y+Z)/C

    因此,

    (Y-Z)/A(B-C)=(Z-X)/B(C-A)

    由轮换对称性,知

    Y-Z/A(B-C)=Z-X/B(C-A)=X-Y/C(A-B)

    2、

    [X/(Y+Z) + Y/(Z+X) + Z/(X+Y)]X=X

    X^2/(Y+Z)+XY/(Z+X)+XZ/(X+Y)=X (1)

    [X/(Y+Z) + Y/(Z+X) + Z/(X+Y)]Y=Y

    XY/(Y+Z)+Y^2/(Z+X)+YZ/(X+Y)=Y (2)

    [X/(Y+Z) + Y/(Z+X) + Z/(X+Y)]Z=Z

    XZ/(X+Z)+YZ/(Z+X)+Z^2/(X+Y)=Z (3)

    (1)+(2)+(3):

    X^2/(Y+Z)+Y^2/(Z+X)+Z^2/(X+Y)+XY/(Z+X)+YZ/(Z+X)+XZ/(X+Y)+YZ/(X+Y)+XY/(Z+X)+XZ/(Z+X)=X+Y+Z

    X^2/(Y+Z)+Y^2/(Z+X)+Z^2/(X+Y)+Y(X+Z)/(Z+X)+Z(X+Y)/(X+Y)+X(Y+Z)/(Z+X)=X+Y+Z

    X^2/(Y+Z)+Y^2/(Z+X)+Z^2/(X+Y)+Y+Z+X=X+Y+Z

    因此:

    X^2/(Y+Z)+Y^2/(Z+X)+Z^2/(X+Y)=0