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Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
=>\(\frac{1}{z}=-\left(\frac{1}{x}+\frac{1}{y}\right)\)
=>\(\left(\frac{1}{z}\right)^3=\left[-\left(\frac{1}{x}+\frac{1}{y}\right)\right]^3\)
=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}+\frac{1}{y}\right)^3\right]\)
=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}\right)^3+3.\left(\frac{1}{x}\right)^2.\frac{1}{y}+3.\frac{1}{x}.\left(\frac{1}{y}\right)^2+\left(\frac{1}{y}\right)^3\right]\)
=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{x}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{y}+\frac{1}{y^3}\right]\)
=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{y^3}\right]\)
=>\(\frac{1}{z^3}=-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left\{-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\right\}\)
\(\frac{1}{z^3}-\left[-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)\right]=-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\)
Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=-3.\frac{1}{x}.\frac{1}{y}.\left(-\frac{1}{z}\right)\)
=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=3.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}\)
=>\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=3.\frac{1}{xyz}\)
=>\(xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=3\)
=>A=3
Vậy A=3
Với x ; y > 0 , cần c/m : \(x^3+y^3\ge xy\left(x+y\right)\)
Ta có : \(x^3+y^3-xy\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-xy\right)=\left(x+y\right)\left(x-y\right)^2\ge0\)
( điều này luôn đúng với mọi x ; y > 0 )
=> BĐT được c/m
Áp dụng vào bài toán , ta có :
\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{xz\left(x+z\right)+xyz}=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{xz\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=\frac{1}{xyz}\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=z;x,y,z>0\)
Xét (1/x+1/y+1/z)^2=1/x^2+1/y^2+1/z^2+2/xy+2/yz+2/xz
=P+2/xy+2/yz+2/xz=P+(2z+2x+2y)/xyz=P+2(x+y+z)/x+y+z=P+2
mà (1/x+1/y+1/z)^2=3
=>p=3-2=1
ta có : a+b+c=0 => (a+b+c)(a2+b2+c2-ab-ac-bc)=0 (đoạn này bạn tự nhân ra rồi rút gọn nhé)
=> a3+b3+c3-3abc=0 => a3+b3+c3= 3abc
thay a=\(\frac{1}{x}\);b=\(\frac{1}{y}\);c=\(\frac{1}{z}\)
=>\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\)=3.\(\frac{1}{xyz}\)
A=xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\)) = xyz .3 . \(\frac{1}{xyz}\)=3
TC \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\left(do\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\right)\)
thay vào \(xyz.\frac{3}{xyz}=3\)
Lời giải:
Đặt $\frac{1}{x}=a; \frac{1}{y}=b; \frac{1}{z}=c$ thì bài toán trở thành:
Cho $a+b+c=0$. Tính $\frac{a^3+b^3+c^3}{abc}$
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Ta có:
$a+b+c=0\Rightarrow a+b=-c$. Khi đó:
$\frac{a^3+b^3+c^3}{abc}=\frac{(a+b)^3-3ab(a+b)+c^3}{abc}$
$=\frac{(-c)^3-3ab(-c)+c^3}{abc}=\frac{-c^3+3abc+c^3}{abc}=\frac{3abc}{abc}=3$
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