Toán Lớp 8: Cho x+y+z=1
giá trị biểu thức S = $\frac{(xy+z)(yz+x)(zx+y)}{(1-x)^2(1-y)^2(1-z)^2}$
help pls
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Toán Lớp 8: Cho x+y+z=1
giá trị biểu thức S = $\frac{(xy+z)(yz+x)(zx+y)}{(1-x)^2(1-y)^2(1-z)^2}$
help pls
Comments ( 2 )
$ => xy + z = xy – x – y + 1 = (x – 1)(y – 1)$
Tương tự:
$ zx + y = (z – 1)(x – 1)$
Thay vào $ => S = 1$