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\(\hept{\begin{cases}x^3-3x-2=2-y\\y^3-3y-2=4-2z\\z^3-3z-2=6-3x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^3-x-2x-2=2-y\\y^3-y-2y-2=2\left(2-z\right)\\z^3-z-2z-2=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\left(x^2-1\right)-2\left(x+1\right)=2-y\\y\left(y^2-1\right)-2\left(y+1\right)=2\left(2-z\right)\\z\left(z^2-1\right)-2\left(z+1\right)=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left[x\left(x-1\right)-2\right]=2-y\\\left(y+1\right)\left[y\left(y-1\right)-2\right]=2\left(2-z\right)\\\left(z+1\right)\left[z\left(z-1\right)-2\right]=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)\left(x^2-x-2\right)=2-y\\\left(y+1\right)\left(y^2-y-2\right)=2\left(2-z\right)\\\left(z+1\right)\left(z^2-z-2\right)=3\left(2-x\right)\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\left(x-2\right)=2-y\\\left(y+1\right)^2\left(y-2\right)=2\left(2-z\right)\\\left(z+1\right)^2\left(z-2\right)=3\left(2-x\right)\end{cases}}\)
Nhân các vế của 3 phương trình với nhau ta được:
\(\left(x+1\right)^2\left(x-2\right)\left(y+1\right)^2\left(y-2\right)\left(z+1\right)^2\left(z-2\right)=6\left(2-y\right)\left(2-z\right)\left(2-x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2=-6\left(y-2\right)\left(z-2\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\left(y-2\right)\left(x-2\right)\left(z-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\left[\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\right]=0\)
Vì \(\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6>0\)
Nên \(\left(x-2\right)\left(y-2\right)\left(z-2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\y-2=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=2\\z=2\end{cases}}}\)
Vậy x = y = z = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\hept{\begin{cases}\left(x+1\right)^2\left(x-2\right)=2-y\\\left(y+1\right)^2\left(y-2\right)=2\left(2-z\right)\\\left(z+1\right)^2\left(z-2\right)=3\left(2-x\right)\end{cases}}\)
nhân từng vế của pt , ta có \(\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2\left(x-2\right)\left(y-2\right)\left(z-2\right)=6\left(2-x\right)\left(2-y\right)\left(2-z\right)\)
\(\Leftrightarrow\left[\left(x+1\right)^2\left(y+1\right)^2\left(z+1\right)^2+6\right]\left(x-2\right)\left(y-2\right)\left(z-2\right)=0\)
đến đây thì dễ rồi, sẽ => x=2, hoặc y=2 hoặc z=2, thay vao rồi giải nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
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\(A=x-2y+3z\left(x,y,z>0\right)\)
\(\left\{{}\begin{matrix}2x+4x+3z=8\left(1\right)\\3x+y-3z=2\left(2\right)\end{matrix}\right.\)
(1) <=> \(5x+5y=10\) <=> x+ y = 2
=> y = 2-x
Từ (1) => \(2x+4\left(2-x\right)+3z=8\)
=> -2x +3z =0
=> \(x=\dfrac{3}{2}z\) => \(z=\dfrac{2}{3}x\) thay vào A
=> \(A=x-2\left(2-x\right)+3.\dfrac{2}{3}x=5x-4\ge-4\)
Vậy Amin = -4.