設(shè):x+y+z=xyz,證明:x(1-y^2)(1-^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz

熱心網(wǎng)友

設(shè):x+y+z=xyz,證明:x(1-y^2)(1-^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz 解：將左邊展開，利用條件x+y+z=xyz，將等式左邊化簡(jiǎn)成右邊．證 因?yàn)閤+y+z=xyz，所以左邊=x(1-z^2-y^2-y^2z^2)+y(1-z^2-x^2+x^2z^2)+(1-y^2-x^2+x^2y^2) =(x+y+z)-xz^2-xy^2+xy^2z^2-yz^2+yx^2+yx^2z^2-zy^2-zx^2+zx^2y^2=xyz-xy(y+x)-xz(x+z)-yz(y+z)+xyz(xy+yz+zx)=xyz-xy(xyz-z)-xz(xyz-y)-yz(xyz-x)+xyz(xy+yz+zx)=xyz+xyz+xyz+xyz=4xyz=右邊．

熱心網(wǎng)友

設(shè):x+y+z=xyz,證明:x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz 因?yàn)閤+y+z=xyz　，所以設(shè)x=tanα 、y=tanβ　、z=tanγ　（其中α+β+γ＝π)因?yàn)?α + 2β + 2γ＝2π所以tan2α + tan2β + tan2γ ＝tan2α*tan2β*tan2γ因?yàn)閠an2α = 2tana/[1-(tanα)^2] ,tan2β = 2tanβ/[1-(tanβ)^2] , 　　tan2γ = 2tanγ/[1-(tanγ)^2]所以(2tana)/[1-(tanα)^2]+ (2tanβ)/[1-(tanβ)^2]+ (2tanγ)/[1-(tanγ)^2]　　＝(2tana)/[1-(tanα)^2] * (2tanβ)/[1-(tanβ)^2]* (2tanγ)/[1-(tanγ)^2]即(2x)/(1-x^2)+(2y)/(1-y^2)+(2z)/(1-z^2)=(8xyz)/(1-x^2)(1-y^2)(1-z^2)所以去分母得：x(1-y^2)(1-z^2)+y(1-x^2)(1-z^2)+z(1-x^2)(1-y^2)=4xyz 。