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\(\frac{x^2-yz}{yz}+1+\frac{y^2-zx}{zx}+1+\frac{z^2-xy}{xy}+1=3\Leftrightarrow\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\)
\(\Leftrightarrow\frac{1}{xyz}\left(x^3+y^3+z^3\right)=3\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\)
Tới đây bạn thay vào nhé :)
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=0\Rightarrow\frac{x+y+z}{xyz}=0\Rightarrow x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(N=\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=\frac{x^3+y^3+z^3}{xyz}=\frac{3xyz}{xyz}=3\)
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]
\(\Rightarrow P=xyz\left[\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)+3xyz\right]\)
\(\Rightarrow P=xyz.3.\frac{1}{xyz}=3\)
\(xyz\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0\\ \Rightarrow yz+xz+xy=0\)
\(A=\frac{xy}{z^2}+\frac{xz}{y^2}+\frac{yz}{x^2}\\ \Leftrightarrow A=\frac{x^3y^3+x^3z^3+y^3z^3}{x^2y^2z^2}\)
Ta có :\(yz+xz+xy=0\)
\(\Rightarrow y^3x^3+x^3z^3+x^3y^3=-3xyz\left(y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz\right)\)
\(=-3xyz\left(yz+xz\right)\left(xz+xy\right)\left(yz+xy\right)\)
\(=-3xyz\left(-xy\right)\left(-yz\right)\left(-xz\right)\\ =3x^2y^2z^2\)
\(\Rightarrow A=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
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 =))
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)nhân lần lượt với x; y; z, ta có:
\(1+\frac{x}{y}+\frac{x}{z}=0\)(1)
\(1+\frac{y}{z}+\frac{y}{x}=0\)(2)
\(1+\frac{z}{x}+\frac{z}{y}=0\)(3)
Từ: (1); (2) và (3) => \(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}=-3\)(*)
Mặt khác: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)quy đồng ta có:
\(\frac{\left(xy+yz+zx\right)}{xyz}=0\)hay xy + yz + zx = 0
Hay: \(\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right).\left(xy+yz+zx\right)=0\)
Khai triển, ta có:
\(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{x}+\frac{y}{x}+\frac{z}{y}=0\)
Vậy: \(\frac{yz}{x^2}+\frac{zx}{y^2}+\frac{xy}{z^2}=-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right)=3\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow xy+yz+xz=0\) (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)
\(\Rightarrow\frac{yz}{x^2+2yz}+\frac{xz}{y^2+2xz}+\frac{xy}{z^2+2xy}=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(y-x\right)\left(y-z\right)}+\frac{xy}{\left(z-x\right)\left(z-y\right)}=1\)