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
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