Ta thấy rằng trong bài này nên áp dụng HĐT

Nếu a+b+c = 0 thì a³ + b³ + c³ = 3abc

Theo bài ra , ta có :

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

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

Ta có :

\(A=\frac{yz}{x^2}+\frac{xz}{y^2}+\frac{xy}{z^2}=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}\)

\(\Leftrightarrow A=xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=xyz.\frac{3}{xyz}=3\)(Vì \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\))

Vậy A = 3

Chúc bạn hok tốt =))

3

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

=>\(\frac{yz+zx+xy}{xyz}=0\Rightarrow xy+yz+zx=0\)

\(P=\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=\frac{y^2z^2yz+z^2x^2xz+x^2y^2xy}{x^2y^2z^2}=\frac{\left(yz\right)^3+\left(zx\right)^3+\left(xy\right)^3}{x^2y^2z^2}\)

Ta có: nếu a+b+c=0 thì a^3 +b^3 +c^3 =3abc

Mà xy+yz+zx=0

=>\(\left(xy\right)^3+\left(yz\right)^3+\left(zx\right)^3=3\cdot xy\cdot yz\cdot zx=3x^2y^2z^2\)

=>\(P=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)

Cho mik sưa chút

\(P=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)

Áp dụng hằng đẳng thức a³ + b³ + c³ = [(a + b + c)(a² + b²+ c²-ab-bc-ca)+3abc]

\(\Rightarrow P=xyz\left[\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)+3xyz\right]\)

\(\Rightarrow P=xyz.3.\frac{1}{xyz}=3\)

\(P=\frac{xyz}{x^3}+\frac{xyz}{y^3}+\frac{xyz}{z^3}=xyz\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)\)

Áp dụng hằng đẳng thức nếu a+b+c=0 thì a³+b³+c³=3abc

Ta có:

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Vậy \(P=xyz.\frac{3}{xyz}=3\)

\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)

\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)

Ta có :\(yz+xz+xy=0\)

         \(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)

                                                  \(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)

                                                   \(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)

      \(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\) (nhân 2 vế với\(xyz\ne0\))

=> x² + 2yz = x² + 2yz - xy - yz - xz = x² - xz - xy + yz = x(x - z) - y(x - z) = (x - y)(x - z).

Tương tự,y² + 2xz = (y - x)(y - z) ; z² + 2xy = (z - x)(z - y)

\(\Rightarrow\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}=1\)

ngu quá có thế cx k làm đc.

sao bạn ns vậy