Toán Lớp 8: Chứng minh đẳng thức sau:
$\frac{1}{x(x-y)(x-z)}+$ $\frac{1}{y(y-z)(y-x)}+$ $\frac{1}{z(z-x)(z-y)}=$ $\frac{1}{xyz}$
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Toán Lớp 8: Chứng minh đẳng thức sau:
$\frac{1}{x(x-y)(x-z)}+$ $\frac{1}{y(y-z)(y-x)}+$ $\frac{1}{z(z-x)(z-y)}=$ $\frac{1}{xyz}$
Comments ( 1 )
1/(x(x-y)(x-z)) +1/(y(y-z)(y-x)) + 1/(z(z-x)(z-y))
= 1/(x(x-y)(x-z)) – 1/(y(y-z)(x-y)) + 1/(z(x-z)(y-z))
=(yz (y-z) – xz (x-z) + xy (x-y))/(xyz (x-y)(y-z)(x-z))
=(yz (y-z) – x^2z +xz^2 + x^2y – xy^2)/(xyz (x-y)(y-z)(x-z))
=(yz(y-z) +x^2 (y-z) – x (y-z)(y+z))/(xyz (x-y)(y-z)(x-z))
=((y-z)(yz +x^2 – xy -xz))/(xyz (x-y)(y-z)(x-z))
=((y-z)(-y (x-z) + x (x-z)))/(xyz (x-y)(y-z)(x-z))
=((x-y)(y-z)(x-z))/(xyz (x-y)(y-z)(x-z))
= 1/(xyz)
-> đpcm