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\(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) \(\dfrac{\Rightarrow1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=a+b+c=0\)
cơ bản \(\left(a+b+c\right)=0\Rightarrow a^3+b^3+c^3=3abc\)
\(\Rightarrow x.y.z\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{1}{abc}.\left(a^3+b^3+c^3\right)=\dfrac{1}{abc}\left(3abc\right)=3=>dpcm\Leftrightarrow dccm\)
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: \(=\dfrac{1}{x\left(x-y\right)\left(x-z\right)}-\dfrac{1}{y\left(y-z\right)\left(x-y\right)}+\dfrac{1}{z\left(x-z\right)\left(y-z\right)}\)
\(=\dfrac{zy\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{zy^2-z^2y-x^2z+xz^2+xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
\(=\dfrac{1}{xyz}\)
Đầu tiên ta cm:\(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+\left(-a-b\right)^3=3abc\)
\(\Leftrightarrow a^3+b^3-a^3-3a^2b-3ab^2-b^3=3abc\)
\(\Leftrightarrow-3a^2b-3ab^2=3abc\)
\(\Leftrightarrow-3ab\left(a+b\right)=3abc\)
\(\Leftrightarrow-3ab\cdot\left(-c\right)=3abc\)(đúng)
Áp dụng:\(\Rightarrow xyz\cdot\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\cdot\dfrac{3}{xyz}=3\left(đpcm\right)\)
\(A=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x^2+2xy+y^2\right)-\left(xz+yz\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
\(=0\)
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\(A=\left(\dfrac{x}{y}+1\right)\left(\dfrac{y}{z}+1\right)\left(\dfrac{z}{x}+1\right)\)
\(=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{z+x}{x}\)
\(=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}\)
\(=-1\)
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\(A=\dfrac{1}{y^2+z^2-x^2}+\dfrac{1}{x^2+z^2-y^2}+\dfrac{1}{x^2+y^2-z^2}\)
\(=\dfrac{1}{\left(y+z\right)^2-2yz-x^2}+\dfrac{1}{\left(x+z\right)^2-2xz-y^2}+\dfrac{1}{\left(x+y\right)^2-2xy-z^2}\)
\(=\dfrac{1}{\left(-x\right)^2-2yz-x^2}+\dfrac{1}{\left(-y\right)^2-2xz-y^2}+\dfrac{1}{\left(-z\right)^2-2xy-z^2}\)
\(=-\dfrac{1}{2}\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xz}\right)\)
\(=-\dfrac{1}{2}\times\dfrac{x+y+z}{xyz}\)
\(=0\)
Giả sử bài toán đã có đầu đủ giả thuyết cần thiết rồi. (Thiếu giả thuyết nhá bác).
\(x^3+y^3+z^3\ge\left(\dfrac{x+y}{2}\right)^3+\left(\dfrac{y+z}{2}\right)^3+\left(\dfrac{z+x}{2}\right)^3\)
\(\Leftrightarrow6\left(x^3+y^3+z^3\right)-3\left(xy^2+xz^3+yx^2+yz^2+zx^2+zy^2\right)\ge0\)
Ta có bổ đề:
\(x^3+x^3+y^3\ge3yx^2\)
Thế vô thì bài toán được chứng minh.
1 cách giải khác:
\(bdt\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge\left(x+y\right)^3+\left(y+z\right)^3+\left(x+z\right)^3\)
\(\Leftrightarrow8\left(x^3+y^3+z^3\right)\ge2\left(x^3+y^3+z^3\right)+xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow6\left(x^3+y^3+z^3\right)\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow3\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(y+z\right)\left(y^2-yz+z^2\right)+3\left(x+z\right)\left(x^2-xz+z^2\right)\ge xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)\)
\(\Leftrightarrow3\left(x+y\right)\left(x-y\right)^2+3\left(y+z\right)\left(y-z\right)^2+3\left(x+z\right)\left(x-z\right)^2=0\)
\("="\Leftrightarrow x=y=z\)
Ta có: \(\left(x+y\right)+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)=x^2+y^2+z^2\)
\(\Rightarrow xy+yz+xz=0\Rightarrow\dfrac{xy+yz+xz}{xyz}=0\)
Hay \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{-1}{z}\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(-\dfrac{1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{3}{xy}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)hay \(\dfrac{1}{x^3}-\dfrac{3}{xyz}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)
\(\Rightarrow\dfrac{1}{x^2}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)