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\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)
A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)
bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha
đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)
bài toán thành \(a^3+b^3+c^3-3abc=0\) nha
\(x^2+y^2-z^2=x^2+\left(y-z\right)\left(y+z\right)=x^2-x\left(y-z\right)=x\left(x-y+z\right)=x\left(-y-y\right)=-2xy\)
Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)
Cộng VTV:
\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Bài này ez thôi, làm mãi rồi.
Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
=>\(\dfrac{xy+yz+xz}{xyz}=0\)
=> xy+yz+zx=0
=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)
Ta có: x2+2yz=x2+yz-xy-zx=(x-y)(x-z)
y2+2xz=y2+xz-xy-yz=(x-y)(z-y)
z2+2xy=z2+xy-yz-xz=(x-z)(y-z)
=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) (\(x,y,z\ne0;x\ne y\ne z\)
\(\Leftrightarrow xy+yz+xz=0\)
\(\Leftrightarrow2yz=yz-xy-xz\)
\(\Leftrightarrow x^2+2yz=\left(x-y\right)\left(x-z\right)\)
CMTT : \(\left\{{}\begin{matrix}y^2+2xz=\left(y-z\right)\left(y-x\right)\\z^2+2xy=\left(z-x\right)\left(z-y\right)\end{matrix}\right.\)
\(A=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{y^2z-yz^2-x^2z+xz^2+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{z^2\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(A=\dfrac{z^2-xz-yz+xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{x\left(y-z\right)-z\left(y-z\right)}{\left(x-z\right)\left(y-1\right)}=1\)
Thề, gõ máy mệt gấp đôi viết tay =))
\(A=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)}{xyz}\)
\(=\dfrac{\left(-z\right)^3+z^3-3xy\left(-z\right)}{xyz}=3\)
Ta có: \(x^2+y^2-z^2\)
\(=\left(x+y\right)^2-z^2-2xy\)
\(=\left(x+y+z\right)\left(x+y-z\right)-2xy\)
\(=-2xy\)
Ta có: \(x^2+z^2-y^2\)
\(=\left(x+z\right)^2-y^2-2xz\)
\(=\left(x+y+z\right)\left(x+z-y\right)-2xz\)
\(=-2xz\)
Ta có: \(y^2+z^2-x^2\)
\(=\left(y+z\right)^2-x^2-2yz\)
\(=\left(x+y+z\right)\left(y+z-x\right)-2yz\)
\(=-2yz\)
Ta có: \(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{xz}{x^2+z^2-y^2}+\dfrac{yz}{y^2+z^2-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}\)
\(=\dfrac{1}{-2}+\dfrac{1}{-2}+\dfrac{1}{-2}\)
\(=\dfrac{-3}{2}\)
\(x,y,z\ne0\)
-Ta c/m: -Với \(a+b+c=0\) thì: \(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0.\left(a^2+b^2+c^2-ab-bc-ca\right)=0\left(đpcm\right)\)
-Quay lại bài toán:
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\dfrac{xy+yz+zx}{xyz}=0\Rightarrow xy+yz+zx=0\)
\(A=\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3}{x^2y^2z^2}=\dfrac{y^3z^3+z^3x^3+x^3y^3-3x^2y^2z^2+3x^2y^2z^2}{x^2y^2z^2}=\dfrac{\left(xy+yz+zx\right)\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=\dfrac{0.\left[x^2y^2+y^2z^2+z^2x^2-xyz\left(x+y+z\right)\right]}{x^2y^2z^2}+3=3\)