Chứng minh từ: \(4\left(y-x\right)\left(z-x\right)+4\left(z-y\right)\left(x-y\right)+4\left(x-z\right)\left(y-z\right)=0\) suy ra x = y =z
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HP
6 tháng 2 2021
\(\dfrac{y-z}{\left(x-y\right)\left(x-z\right)}+\dfrac{z-x}{\left(y-z\right)\left(y-x\right)}+\dfrac{x-y}{\left(z-x\right)\left(z-y\right)}\)
\(=\dfrac{z-x+x-y}{\left(x-y\right)\left(z-x\right)}+\dfrac{x-y+y-z}{\left(y-z\right)\left(x-y\right)}+\dfrac{y-z+z-x}{\left(z-x\right)\left(y-z\right)}\)
\(=\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}+\dfrac{1}{x-y}+\dfrac{1}{z-x}+\dfrac{1}{y-z}\)
\(=\dfrac{2}{x-y}+\dfrac{2}{y-z}+\dfrac{2}{z-x}\)
26 tháng 11 2022
a: \(\dfrac{y}{\left(x-y\right)\left(y-z\right)}-\dfrac{z}{\left(y-z\right)\left(x-z\right)}-\dfrac{x}{\left(x-y\right)\left(x-z\right)}\)
\(=\dfrac{xy-yz-xz+yz-xy+xz}{\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
=0
c: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
9 tháng 9 2020
Bài dễ mừ, có phải Croatia thật ko vậy :)) (viết đề bị nhầm, là x,y,z dương chứ :))
Áp dụng Cauchy-Schwarz dạng cộng mẫu số:
\(\frac{x^2}{\left(x+y\right)\left(x+z\right)}+\frac{y^2}{\left(y+z\right)\left(y+x\right)}+\frac{z^2}{\left(z+x\right)\left(z+y\right)}\ge\)
\(\frac{\left(x+y+z\right)^2}{\left(x+y\right)\left(x+z\right)+\left(y+z\right)\left(y+x\right)+\left(z+x\right)\left(z+y\right)}=\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\)
\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}\)
Xét \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\Rightarrow\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}\)
\(=\frac{\left(x+y+z\right)^2}{\frac{4}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
Dấu bằng xảy ra khi và chỉ khi x=y=z, Xong! :))
\(4\left(y-z\right)\left(z-x\right)+4\left(z-y\right)\left(x-y\right)+4\left(x-z\right)\left(y-z\right)=0\)
\(\Leftrightarrow4yz-4xy-4xz+4x^2+4xz-4yz-4xy+4y^2+4xy-4xz-4yz+4z^2=0\)\(\Leftrightarrow4x^2-4xy-4xz+4y^2-4yz+4z^2=0\)
\(\Leftrightarrow4\left(x^2-xy-xz+y^2-yz+z^2\right)=0\)
\(\Leftrightarrow x^2+y^2+z^2-xy-xz-yz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)
\(\Leftrightarrow x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\\left(x-z\right)^2=0\\\left(y-z\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-z=0\\y-z=0\end{matrix}\right.\Leftrightarrow x=y=z\)