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![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\left|\frac{1}{2}-x\right|+\left|x-\frac{1}{4}\right|+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\)
\(VT\ge\left|\frac{1}{2}-x+x-\frac{1}{4}\right|+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\)
\(VT\ge\frac{1}{4}+\left|x-\frac{1}{3}\right|+\left|y-\frac{1}{5}\right|\ge\frac{1}{4}\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-\frac{1}{3}=0\\y-\frac{1}{5}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\y=\frac{1}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(y=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
hình như mk thấy có phần tương tự trong sbt oán 7 ở phần nào đó thì phải . Bạn về nhà tìm thử xem sau đó mở đáp án ở sau mà coi
Lí luận chung cho cả 3 câu :
Vì GTTĐ luôn lớn hơn hoặc bằng 0
a) \(\Rightarrow\hept{\begin{cases}x+\frac{3}{7}=0\\y-\frac{4}{9}=0\\z+\frac{5}{11}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{-3}{7}\\y=\frac{4}{9}\\z=\frac{-5}{11}\end{cases}}}\)
b)\(\Rightarrow\hept{\begin{cases}x-\frac{2}{5}=0\\x+y-\frac{1}{2}=0\\y-z+\frac{3}{5}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{10}\\z=\frac{7}{10}\end{cases}}}\)
c)\(\Rightarrow\hept{\begin{cases}x+y-2,8=0\\y+z+4=0\\z+x-1,4=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=2,8\\y+z=-4\\z+x=1,4\end{cases}}}\)
\(\Rightarrow x+y+y+z+z+x=2,8-4+1,4\)
\(\Rightarrow2\left(x+y+z\right)=0,2\)
\(\Rightarrow x+y+z=0,1\)
Từ đây tìm đc x, y, z
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì: \(Ix+\frac{1}{2}I\ge0\)
\(Iy-\frac{3}{4}I\ge0\)
\(Iz-1I\ge0\)
Mà \(Ix+\frac{1}{2}I+Iy-\frac{3}{4}I+Iz-1I=0\)
=> \(x+\frac{1}{2}=0\) và \(y-\frac{3}{4}=0\) và \(z-1=0\)
<=> \(x=-\frac{1}{2}\) và \(y=\frac{3}{4}\) và \(z=1\)
Vậy \(x=-\frac{1}{2}\) và \(y=\frac{3}{4}\) và \(z=1\)
phần B lm tương tự nha
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
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 \(x=\frac{25-25y}{18}\)
\(y=\frac{25}{6}-3x\)
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