Nếu x . 0,(72) = 2 . (96)y . Vậy x/y=...........
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ne_-^+y;y\ne0\)
Từ PT thứ 2 ta có:\(\dfrac{96}{x-y}+\dfrac{72}{x-y}=\dfrac{24}{y}\)
<=>\(\dfrac{168}{x-y}=\dfrac{24}{y}\)
<=>\(\dfrac{168}{x-y}=\dfrac{168}{7y}\)
<=>x-y=7y
<=>x=8y
Thay x=8y vào PT thứ nhất:
\(\dfrac{96}{8y+y}+\dfrac{96}{8y-y}=14\)
<=>\(\dfrac{32}{3y}+\dfrac{96}{7y}=14\)
<=>32.7y+96.3y=294y2
<=>512y=294y2
<=>y=\(\dfrac{256}{147}\left(Doy\ne0\right)\)
=>x=8y=\(\dfrac{2048}{147}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
(1)+(3)-(2) \(\Rightarrow x\left(x+y+z\right)=24\) (4)
\(\left(1\right)+\left(2\right)-\left(3\right)\Rightarrow y\left(x+y+z\right)=48\) (5)
\(\left(2\right)+\left(3\right)-\left(1\right)\Rightarrow z\left(x+y+z\right)=72\) (6)
Cộng vế với vế: \(\Rightarrow\left(x+y+z\right)^2=144\Rightarrow\left[{}\begin{matrix}x+y+z=12\\x+y+z=-12\end{matrix}\right.\)
- Với \(x+y+z=12\) (7) lần lượt chia vế cho vế cho (4); (5); (6) cho (7)
- Với \(x+y+z=-12\) (8) lần lượt chia vế cho vế của (4); (5); (6) cho (8)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ \(\left\{{}\begin{matrix}\left(x-2\right)^{72}\ge0\\\left(y+1\right)^{70}\ge0\end{matrix}\right.\)
Mà \(\left(x-2\right)^{72}+\left(y+1\right)^{70}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^{72}=0\\\left(y+1\right)^{70}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Vậy ...
2/ \(\left\{{}\begin{matrix}\left|x+1\right|\ge0\\\left|y-3\right|\ge0\end{matrix}\right.\)
Mà \(\left|x+1\right|+\left|y-3\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=0\\\left|y-3\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\)
Vậy ...
3/ \(\left\{{}\begin{matrix}\left(2x-10\right)^{100}\ge0\\\left(x-y\right)^{102}\ge0\end{matrix}\right.\)
Mà \(\left(2x-10\right)^{100}+\left(x-y\right)^{102}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-10\right)^{100}=0\\\left(x-y\right)^{102}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-10=0\\x-y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=5\end{matrix}\right.\)
Vậy ....
4/ \(\left\{{}\begin{matrix}\left|2x+8\right|\ge0\\\left|y+x\right|\ge0\end{matrix}\right.\)
Mà \(\left|2x+8\right|+\left|y+x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|2x+8\right|=0\\\left|y+x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+8=0\\y+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-8\\y=8\end{matrix}\right.\)
Vậy ..