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13 tháng 4 2020

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18 tháng 4 2020

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(x+y+z)^2=x^2+y^2+z^2

=>2(xy+yz+xz)=0

=>xy+xz+yz=0

=>xy/xyz+xz/xyz+yz/xyz=0

=>1/x+1/y+1/z=0

14 tháng 9 2018

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=0\)

=> x + y + z = 0

Lai co: x3 + y3 +z3 - 3xyz = (x+y+z).(x2+y2+z2 - xy - yz - zx)

             x3 + y3 + z3 - 3xyz = 0

=> x3 + y3 + z3 = 3xyz

14 tháng 9 2018

ta co: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}.\)

=> 1/xy + 1/yz + 1/xz = 0

=> x + y + z = 0

Lai co: x3 + y3 +z3 - 3xyz = (x+y+z).(x2+y2+z2 - xy - yz - zx)

             x3 + y3 + z3 - 3xyz = 0

=> x3 + y3 + z3 = 3xyz

13 tháng 12 2016

CÓ:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-\frac{1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}-\frac{3}{xyz}=-\frac{1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

\(A=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz\cdot\frac{3}{xyz}=3\)