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Theo đầu bài ta có:
\(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}\)
\(\Rightarrow\frac{2\cdot\left(x+1\right)}{2\cdot2}=\frac{3\cdot\left(y+3\right)}{3\cdot4}=\frac{4\cdot\left(z+5\right)}{4\cdot6}\)
\(\Rightarrow\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}\)
\(=\frac{\left(2x+2\right)+\left(3y+9\right)+\left(4z+20\right)}{4+12+24}\)
\(=\frac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{4+12+24}\)
\(=\frac{9+31}{40}=1\)
\(\Rightarrow\hept{\begin{cases}x=1\cdot2-1=1\\y=1\cdot4-3=1\\z=1\cdot6-5=1\end{cases}}\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{-2.\left(-4k\right)+\left(-7k\right)+5.3k}{-4k-3.\left(-7k\right)-6.3k}=\dfrac{16k}{-1k}=-16\)
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...
Ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) => \(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{2z^2}{32}=\frac{x^2+y^2-2z^2}{4+9-32}=\frac{76}{-19}=-4\)
=> \(\hept{\begin{cases}\frac{x^2}{4}=-4\\\frac{y^2}{9}=-4\\\frac{2z^2}{32}=-4\end{cases}}\) => \(\hept{\begin{cases}x^2=-4.4=-16\\y^2=-4.9=-36\\z^2=\left(-4.32\right):2=-64\end{cases}}\) => ko có giá trị x,y,z thõa mãn
Ta có: \(-2x=5y\) => \(\frac{x}{5}=\frac{y}{-2}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{5}=\frac{y}{-2}=\frac{x+y}{5-2}=\frac{30}{3}=10\)
=> \(\hept{\begin{cases}\frac{x}{5}=10\\\frac{y}{-2}=10\end{cases}}\) => \(\hept{\begin{cases}x=10.5=50\\y=10.\left(-2\right)=-20\end{cases}}\)
Vậy ..
\(\frac{x}{-3}=\frac{y}{-7}\Rightarrow\frac{2x}{-6}=\frac{4y}{-28}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2x}{-6}=\frac{4y}{-28}=\frac{2x+4y}{(-6)+(-28)}=\frac{68}{-34}=-2\)
Vậy : \(\hept{\begin{cases}\frac{x}{-3}=-2\\\frac{y}{-7}=-2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=6\\y=14\end{cases}}\)
a) Vì \(\left|2x+4\right|\ge0;\left|y\right|\ge0\)
mà \(\left|2x+4\right|+\left|y\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|2x+4\right|=0\\\left|y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-2;0\right)\)
dạ cho mình hỏi bạn biết làm câu b không ạ