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Ta có: x + y + z = 0 <=> y + z = -x
(y+z)5 = (-x)5
y5 + z5 + 5y4z + 10y3z2 + 10y2z3 + 5yz4 = -x5
y5 + z5 + 5y4z + 10y3z2 + 10y2z3 + 5yz4 + x5 = 0
x5 + y5 + z5 +5xyz[ y3 + 2y2z + 2yz2 + z3 ] = 0
x5 + y5 + z5 + 5xyz[(y+z)(y2 -yz -z2)+ 2yz(x+z)] = 0
x5 + y5 + z5 +5xyz[(y+z)(y2 +yz + z2)] = 0
2.(x5 + y5 + z5) + 5xyz(y+z)(y2+yz+z2) - (x5 + y5 + z5) = 0
2(x5 + y5 + z5) - 5xyz[(y2+2yz+z2)+y2+z2] = 0
2(x5 + y5 + z5) = 5xyz[(y+z)2 + y2 + z2]
2(x5 + y5 + z5) = 5xyz[(-x)2 + y2 + z2]
2(x5 + y5 + z5) = 5xyz(x2 + y2 + z2).
\(y+z=-x\)
\(\left(y+z\right)^5=-x^5\)
\(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(x^5+y^5+z^5+5yz\left[\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right]=0\)
\(x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)-5xyz\left(\left(y^2+2yz+z^2\right)+y^2+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
a, Ta có: \(2\left(x^8+y^8\right)\ge\left(x^3+y^3\right)\left(x^5+y^5\right)\)
\(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)
Ta CM: \(\Leftrightarrow x^8+y^8\ge x^5y^3+x^3y^5\)
Áp dụng bđt Cô si:
\(x^8+x^8+x^8+x^8+x^8+y^8+y^8+y^8\ge8x^5y^3\) (*)
Tương tự, \(5y^3+3x^3\ge8x^3y^5\) (**)
Từ (*), (**) \(\Rightarrowđpcm\)
\(y+z=-x\)
\(\left(y+z\right)^5=-x^5\)
\(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(x^5+y^5+z^5+5yz\left(\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right)=0\)
\(x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)-5xyz\left(\left(y^2+2yz+z^2\right)+y^2+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
Ta có: \(y+z=-x\)
\(\left(y+z\right)^5=-x^5\)
\(y^5+5y^4z+10y^3z^2+10y^2z^3+5yz^4+z^5+x^5=0\)
\(x^5+y^5+z^5+5yz\left(y^3+2y^2z+2yz^2+z^3\right)=0\)
\(x^5+y^5+z^5+5yz\left(\left(y+z\right)\left(y^2-yz+z^2\right)+2yz\left(y+z\right)\right)=0\)
\(x^5+y^5+z^5+5yz\left(y+z\right)\left(y^2+yz+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)-5xyz\left(\left(y^2+2yz+z^2\right)+y^2+z^2\right)=0\)
\(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
\(\left\{{}\begin{matrix}3x^2+xz-yz+y^2=2\left(1\right)\\y^2+xy-yz+z^2=0\left(2\right)\\x^2-xy-xz-z^2=2\left(3\right)\end{matrix}\right.\)
Lấy (2) cộng (3) ta được
\(x^2+y^2-yz-zx=2\) (4)
Lấy (1) - (4) ta được
\(2x\left(x+z\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-z\end{matrix}\right.\)
Xét 2 TH rồi thay vào tìm được y và z
1. \(\left\{{}\begin{matrix}6xy=5\left(x+y\right)\\3yz=2\left(y+z\right)\\7zx=10\left(z+x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x+y}{xy}=\dfrac{6}{5}\\\dfrac{y+z}{yz}=\dfrac{3}{2}\\\dfrac{z+x}{zx}=\dfrac{7}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{5}\\\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{3}{2}\\\dfrac{1}{z}+\dfrac{1}{x}=\dfrac{7}{10}\end{matrix}\right.\)
Đến đây thì dễ rồi nhé
2[x5x3-4x-9y-8z]x[4x-4x+6xy]=0
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