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a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)
\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)
mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)
\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
b) Tương tự câu a, ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
c. Tương tự, ta có:
\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)
a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...
b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...
c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...
|x|=-2/3 là vô lý rồi bạn nên cái này không cần xét trường hợp luôn.
Mà nó sẽ ra ngay là C không có giá trị
\(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}=\frac{x+y-z}{4+5-6}=\frac{x+y-z}{3}\Rightarrow x+y-z=\frac{3x}{4}\)
\(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}=\frac{x}{4}=\frac{2y}{10}=\frac{z}{6}=\frac{x+2y-z}{4+10-6}=\frac{x+2y-z}{8}\Rightarrow x+2y-z=\frac{8x}{4}=2x\)
\(B=\frac{x+y-z}{x+2y-z}=\frac{\frac{3x}{4}}{2x}=\frac{3x}{4.2x}=\frac{3}{8}\)
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{2}\) và \(x-3y=20\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{5}=\dfrac{3y}{9}=\dfrac{z}{2}=\dfrac{x-3y}{5-9}=\dfrac{20}{-4}=-5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-5< =>x=-25\\\dfrac{y}{3}=-5< =>y=-15\\\dfrac{z}{2}=-5< =>z=-10\end{matrix}\right.\)
Vậy ....
TH1:\(\hept{\begin{cases}x+\frac{1}{2}>0\\x-\frac{1}{3}>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-\frac{1}{2}\\x>\frac{1}{3}\end{cases}\Leftrightarrow}x>\frac{1}{3}}\)
TH2:\(\hept{\begin{cases}x+\frac{1}{2}< 0\\x-\frac{1}{3}< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -\frac{1}{2}\\x< \frac{1}{3}\end{cases}\Leftrightarrow}x< -\frac{1}{2}}\)
vậy để biểu thức \(\left(x+\frac{1}{2}\right)\left(x-\frac{1}{3}\right)>0\)thì x > 1/3 hoặc x < (-1/2)
\(\left|x-\dfrac{2}{5}\right|=1\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=1\\\dfrac{2}{5}-x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Với \(x=\dfrac{7}{5};y=3\Rightarrow B=\left|\dfrac{7}{5}-\dfrac{3}{8}\right|+\dfrac{1}{3}\cdot3=\dfrac{41}{40}+1=\dfrac{81}{40}\)
Với \(x=-\dfrac{3}{5};y=3\Rightarrow B=\left|-\dfrac{3}{5}-\dfrac{3}{8}\right|+\dfrac{1}{3}\cdot3=\dfrac{39}{40}+1=\dfrac{79}{40}\)