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a: Ta có: \(\left(x+y\right)^2\)
\(=x^2+2xy+y^2\)
\(\Leftrightarrow x^2+y^2=\dfrac{\left(x+y\right)^2}{2xy}\ge\dfrac{\left(x+y\right)^2}{2}\forall x,y>0\)
Xin lỗi mk viết nhầm
(x+y+z)2-x2-y2-z2 =x2+y2+z2+2(xy+yz+xz)-x2-y2-z2
(x+y+z)2-x2-y2-z2
=x2+y2+2(xy+yz+xz)-x2-y2-z2
= 2(xy+yz+xz)
Vậy hằng đẳng thức được chứng minh
a)(x-y)3+(y-z)3+(z-x)3
=3(x-y+y-z+z-x)=3
b)nhân vào là rồi đối trừ là hết luôn ( nhưng là mũ 2 hay nhân 2 v mk là theo nhân 2 nhé]
Sửa đề \(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx\)(hằng đẳng thức cho 3 số )
\(\Rightarrow\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\left(đpcm\right)\)
Vậy
Ta có:
VT= \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=x^2+y^2+z^2+2xy+2yz+2zx-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\) = VP
=> đpcm
\(\left(x+y+z\right)^2-x^2-y^2-z^2=2\left(xy+yz+zx\right)\)
Biến đổi vế trái:
VT\(\)\(\)\(=\left[\left(x+y\right)+z\right]^2-x^2-y^2-z^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2-x^2-y^2-z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2\)\
\(=2xy+2yz+2zx\)
\(=2\left(xy+yz+zx\right)=\) VP
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\left\{{}\begin{matrix}1+\dfrac{x}{y}+\dfrac{x}{z}=0\\\dfrac{y}{x}+1+\dfrac{y}{z}=0\\\dfrac{z}{x}+\dfrac{z}{y}+1=0\end{matrix}\right.\\ \Rightarrow\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=-3\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\\ \Rightarrow\dfrac{yz+xz+xy}{xyz}=0\\ \Rightarrow yz+xz+xy=0\)
\(\Rightarrow\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\left(xy+xz+yz\right)=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}+\dfrac{x}{y}+\dfrac{x}{z}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{x}+\dfrac{z}{y}=0\\ \Rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{-1}{z}\)
\(\Rightarrow\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^3=\left(\dfrac{-1}{z}\right)^3\)
\(\Leftrightarrow\dfrac{1}{x^3}+3\dfrac{1}{x^2}\dfrac{1}{y}+3\dfrac{1}{x}\dfrac{1}{y^2}+\dfrac{1}{y^3}=\dfrac{-1}{z^3}\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3.\dfrac{1}{x}\dfrac{1}{y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Leftrightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=-3\dfrac{1}{x}\dfrac{1}{y}\dfrac{-1}{z}\)
\(\Leftrightarrow\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)xyz=3\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}.xyz\)
\(\Leftrightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3\)
\(VT=\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left[\left(x+y\right)+z\right]^2-x^2-y^2-z^2\)
\(=\left(x+y\right)^2+2\left(x+y\right)z+z^2-x^2-y^2-z^2\)
\(=x^2+2xy+y^2+2xz+2yz+z^2-x^2-y^2-z^2\)
\(=2xy+2yz+2zx\)
\(=2\left(xy+yz+zx\right)\)
\(=VP\)
Vậy...