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Ta có
\(\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|y+\frac{3}{2}\right|\ge0\\\left|x+y-z-\frac{1}{2}\right|\ge0\end{cases}\)
Maf \(\left|x-\frac{1}{2}\right|+\left|y+\frac{3}{2}\right|+\left|x+y-z-\frac{1}{2}\right|=0\)
\(\Rightarrow\begin{cases}x-\frac{1}{2}=0\\y+\frac{3}{2}=0\\x+y-z-\frac{1}{2}=0\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\x+y-z=\frac{1}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\\frac{1}{2}-\frac{3}{2}-z=\frac{1}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\-z=\frac{3}{2}\end{cases}\)
\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=-\frac{3}{2}\\z=-\frac{3}{2}\end{cases}\)
\(b,\frac{z}{7}=-\frac{11}{-28}\)
\(\Leftrightarrow z.\left(-28\right)=-11.7\)
\(\Leftrightarrow z.\left(-28\right)=-77\)
\(\Leftrightarrow z=\frac{11}{4}\)
\(a,-\frac{2}{3}=\frac{x-3}{-6}=\frac{10}{5-y}=\frac{4-2z}{9}\)
Xét :
\(-\frac{2}{3}=\frac{x-3}{-6}\)
\(\Leftrightarrow-2.\left(-6\right)=\left(x-3\right).3\)
\(\Leftrightarrow12=\left(x-3\right).3\)
\(\Leftrightarrow4=x-3\Leftrightarrow x=7\)
Xét
\(-\frac{2}{3}=\frac{10}{5-y}\)
\(\Leftrightarrow-2.\left(5-y\right)=10.3\)
\(\Leftrightarrow-10+2y=30\)
\(\Leftrightarrow2y=40\Leftrightarrow y=20\)
Xét :
\(-\frac{2}{3}=\frac{4-2z}{9}\)
\(\Leftrightarrow-2.9=\left(4-2z\right).3\)
\(\Leftrightarrow-18=\left(4-2z\right).3\)
\(\Leftrightarrow-6=4-2z\)
\(\Leftrightarrow10=2z\Leftrightarrow z=5\)
Vậy \(\left(x;y;z\right)=\left(7;20;5\right)\)
\(\frac{x}{24}=\frac{-2}{3}\Leftrightarrow x=\frac{-2\times24}{3}=-16\)
\(\frac{y}{-18}=\frac{-2}{3}\Leftrightarrow y=\frac{-2\times-18}{3}=12\)
\(\frac{-28}{z}=\frac{-2}{3}\Leftrightarrow z=\frac{-28\times3}{-2}=42\)
\(\frac{-10}{t}=\frac{-2}{3}\Leftrightarrow t=\frac{-10\times3}{-2}=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\frac{x}{10}=\frac{y}{6}=\frac{z}{21}=\frac{5x}{50}=\frac{y}{6}=\frac{2z}{42}=\frac{5x+y-2z}{50+6-42}=\frac{28}{14}=2\)
\(\Rightarrow\begin{cases}\frac{x}{10}=2\\\frac{y}{6}=2\\\frac{z}{21}=2\end{cases}\)\(\Rightarrow\begin{cases}x=20\\y=12\\z=42\end{cases}\)
Vậy x=20;y=12;z=42
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