Tìm x, y biết:
\(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{3}\right|+\left|x-\frac{1}{4}\right|+\left|y-\frac{1}{5}\right|=\frac{1}{4}\)
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13 tháng 3 2020
phá ngoặc tính BT , nên kết quả sẽ ko ra con số nhận định !!! tui thử thui nha bà !
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)
\(3x+y-\frac{47}{12}=\frac{1}{4}\)
\(3x+y=\frac{25}{6}\)
\(3x=\frac{25}{6}-y\)
\(x=\frac{25-25y}{18}\)
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|y-5\right|+\left|x+\frac{1}{4}\right|=\frac{1}{4}\)
\(x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}=\frac{1}{4}\)
\(3x+y-\frac{47}{12}=\frac{1}{4}\)
\(3x+y=\frac{25}{6}\)
\(y=\frac{25}{6}-3x\)
Vậy \(x=\frac{25-25y}{18}\)
\(y=\frac{25}{6}-3x\)
17 tháng 3 2020
Ta có:
\(|x+\frac{1}{2}|\ge x+\frac{1}{2}\forall x;|x+\frac{1}{3}|\ge x+\frac{1}{3}\forall x;|y-5|\ge y-5\forall y;|x+\frac{1}{4}|\ge x+\frac{1}{4}\forall x\)
\(\Rightarrow|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)
Mà \(|x+\frac{1}{2}|+|x+\frac{1}{3}|+|y-5|+|x+\frac{1}{4}|=\frac{1}{4}\)
\(\Rightarrow\frac{1}{4}\ge x+\frac{1}{2}+x+\frac{1}{3}+y-5+x+\frac{1}{4}\)
\(\Rightarrow\frac{1}{4}\ge3x+y-\frac{47}{12}\)
\(\Rightarrow3x+y\le\frac{25}{6}\)
\(\Rightarrow x\le\frac{\frac{25}{6}-y}{3}\)
Thay vào tính y
NN
20 tháng 5 2020
Đặt \(A=\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{3}\right|+\left|x-\frac{1}{4}\right|+\left|y-\frac{1}{5}\right|=\frac{1}{4}\)
\(\Rightarrow A=\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{4}\right|+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|=\frac{1}{4}\)
Xét \(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{4}\right|\)ta có:
\(\left|x-\frac{1}{2}\right|+\left|x-\frac{1}{4}\right|=\left|x-\frac{1}{2}\right|+\left|\frac{1}{4}-x\right|\ge\left|x-\frac{1}{2}+\frac{1}{4}-x\right|=\left|\frac{-1}{4}\right|=\frac{1}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow\left(x-\frac{1}{2}\right)\left(\frac{1}{4}-x\right)\ge0\)
TH1: \(\hept{\begin{cases}x-\frac{1}{2}\le0\\\frac{1}{4}-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{1}{2}\\\frac{1}{4}\le x\end{cases}}\Leftrightarrow\frac{1}{4}\le x\le\frac{1}{2}\)
TH2: \(\hept{\begin{cases}x-\frac{1}{2}\ge0\\\frac{1}{4}-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\\frac{1}{4}\ge x\end{cases}}\Leftrightarrow\frac{1}{4}\ge x\ge\frac{1}{2}\)( vô lý )
mà \(\left|x-\frac{1}{3}\right|\ge0\forall x\); \(\left|y-\frac{1}{5}\right|\ge0\forall y\)
\(\Rightarrow A\ge\frac{1}{4}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{1}{4}\le x\le\frac{1}{2}\\x-\frac{1}{3}=0\\y-\frac{1}{5}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{1}{4}\le x\le\frac{1}{2}\\x=\frac{1}{3}\\y=\frac{1}{5}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=\frac{1}{5}\end{cases}}\)
Vậy \(x=\frac{1}{3}\)và \(y=\frac{1}{5}\)
19 tháng 7 2021
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Hoc24 đang bị lỗi m ơi :)) m tag t cs nhận đc tb đâu
\(VT=\left|\frac{1}{2}-x\right|+\left|x-\frac{1}{4}\right|+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\)
\(VT\ge\left|\frac{1}{2}-x+x-\frac{1}{4}\right|+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\)
\(VT\ge\frac{1}{4}+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\ge\frac{1}{4}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-\frac{1}{3}=0\\y-\frac{1}{5}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{1}{5}\end{matrix}\right.\)