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\(\left\{{}\begin{matrix}\left(x+1\right)\left(x^2+1\right)=y^3+1\\\left(y+1\right)\left(y^2+1\right)=z^3+1\\\left(z+1\right)\left(z^2+1\right)=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+x^2+x=y^3\left(1\right)\\y^3+y^2+y=z^3\\z^3+z^2+z=x^3\end{matrix}\right.\)
Giả sử \(x>y\Rightarrow x^3+x^2+x>y^3+y^2+y\)
\(\Rightarrow y^3>z^3\Leftrightarrow y>z\left(2\right)\)
\(\Rightarrow y^3+y^2+y>z^3+z^2+z\Rightarrow z>x\left(3\right)\)
Từ \(\left(2\right);\left(3\right)\Rightarrow y>x\) (Vô lí)
Giả sử \(x< y\Rightarrow x^3+x^2+x< y^3+y^2+y\)
\(\Rightarrow y^3< z^3\Leftrightarrow y< z\left(4\right)\)
\(\Rightarrow y^3+y^2+y< z^3+z^2+z\Rightarrow z< x\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow y< x\) (Vô lí)
\(\Rightarrow x=y=z\)
\(\left(1\right)\Leftrightarrow x^3+x^2+x=x^3\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow x=y=z=0\) hoặc \(x=y=z=-1\)
\(\left\{{}\begin{matrix}x+y+z=1\\\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=64\end{matrix}\right.\)
Ta có:
\(1=x+y+z\ge3\sqrt[3]{xyz}\)
\(\Leftrightarrow xyz\le\dfrac{1}{27}\)
Ta có: \(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{1}{xyz}\)
\(\ge1+\dfrac{3}{\sqrt[3]{x^2y^2z^2}}+\dfrac{3}{\sqrt[3]{xyz}}+\dfrac{1}{xyz}\)
\(=1+\dfrac{3}{\sqrt[3]{\dfrac{1}{27^2}}}+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{1}{\dfrac{1}{27}}=64\)
Dấu = xảy ra khi \(x=y=z=\dfrac{1}{3}\)
Phần cuối là:
\(\ge1+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{3}{\sqrt[3]{\dfrac{1}{27}}}+\dfrac{1}{\dfrac{1}{27}}=64\), không phải là dấu ''=''.
\(\Leftrightarrow\left\{{}\begin{matrix}6x-6-2y+4=0\\4x+4-3y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\4x-3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
Giải:
Đặt: (x + y) = a ; (y + z) = b ; (z + x) = c
HPT <=> \(\left\{{}\begin{matrix}ab=187\\bc=154\\ca=238\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{187}{a}\\\dfrac{187}{a}\cdot c=154\\c\cdot a=238\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{187}{a}\\c=\dfrac{154a}{187}\\\dfrac{154a}{187}\cdot a=238\end{matrix}\right.\) => \(154a^2=238\cdot187=44506\)
=> \(a^2=\dfrac{44506}{154}=289\Rightarrow a=\sqrt{289}=17\)
=> b = \(\dfrac{187}{17}=11\) ; c = \(\dfrac{238}{17}=14\)
Hay \(\left\{{}\begin{matrix}x+y=17\\y+z=11\\z+x=14\end{matrix}\right.\)
\(\Rightarrow x+y+y+z+z+x-17+11+14=42\)
\(\Leftrightarrow2\left(x+y+z\right)=42\Rightarrow x+y+z=21\)
=> \(\left\{{}\begin{matrix}x=21-\left(y+z\right)=21-11=10\\y=21-\left(z+x\right)=21-14=7\\z=21-\left(x+y\right)=21-17=4\end{matrix}\right.\)
Vậy ..........................
Đặt x + y = a ( a > 0 )
y + z = b ( b > 0 )
x + z = c (c > )
Khi đó hệ pt thành :
\(\left\{{}\begin{matrix}ab=187\left(1\right)\\bc=154\left(2\right)\\ac=238\left(3\right)\end{matrix}\right.\)
Nhân (1) (2) (3) vế theo vế được: abc = 2618 (4)
Lần lượt chia (4) cho (1) (2) (3) ta được:
\(\left\{{}\begin{matrix}a=17\\b=11\\c=14\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x+y=17\\y+z=11\\x+z=14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-z=6\\x+z=14\end{matrix}\right.\Leftrightarrow x=10\Rightarrow y=7\) và \(z=4\)
Vậy nghiệm của hệ pt là (10;7;4)