Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{5}\right|\ge0\\\left|x+\dfrac{1}{15}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{5}\right|+\left|x+\dfrac{1}{15}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{5}+x+\dfrac{1}{15}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\)
Vậy ..
\(\Leftrightarrow\left[{}\begin{matrix}\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=-4\\\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|-3=4\end{matrix}\right.\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{4}\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{4}=7\\\dfrac{1}{2}x-\dfrac{1}{4}=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{29}{4}\\\dfrac{1}{2}x=-\dfrac{27}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{2}\\x=-\dfrac{27}{2}\end{matrix}\right.\)
\(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| - \(\dfrac{1}{5}\)= \(\dfrac{1}{6}\)
=> \(\dfrac{1}{2}\)| \(\dfrac{1}{3}x\) - \(\dfrac{1}{4}\)| = \(\dfrac{11}{30}\)
=> | \(\dfrac{1}{3}x\)- \(\dfrac{1}{4}\)| = \(\dfrac{11}{15}\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{11}{15}\\\dfrac{1}{3}x-\dfrac{1}{4}=\dfrac{-11}{15}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{59}{60}\\\dfrac{1}{3}x=\dfrac{-29}{60}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{59}{20}\\x=\dfrac{-29}{20}\end{matrix}\right.\)
Chúc bạn học tốt !
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|=4x\)
Ta có:
\(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{6}\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}>0\\x+\dfrac{1}{3}>0\\x+\dfrac{1}{6}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|=x+\dfrac{1}{2}\\\left|x+\dfrac{1}{3}\right|=x+\dfrac{1}{3}\\\left|x+\dfrac{1}{6}\right|=x+\dfrac{1}{6}\end{matrix}\right.\)
Thay vào ta được:
\(x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{6}=4x\)
\(\Rightarrow x=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}=1\)
Vậy...................
Chúc bạn học tốt!!!
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)
\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)
\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)
\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)
Vì \(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
\(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)
\(\Leftrightarrow\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}\)
Theo tính chất của dãy tỉ số bằng nhau, có:
\(\dfrac{12x-8y}{16}=\dfrac{6z-12x}{9}=\dfrac{8y-6z}{4}=\dfrac{12x-8x+6z-12x+8y-6z}{16+9+4}=\dfrac{0}{29}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}12x=8y\\6z=12x\\8y=6z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=\dfrac{y}{12}\\\dfrac{x}{6}=\dfrac{z}{12}\\\dfrac{y}{6}=\dfrac{z}{8}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\\dfrac{x}{2}=\dfrac{z}{4}\\\dfrac{y}{3}=\dfrac{z}{4}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\left(đpcm\right)\)
Kết luận ...
3a)Vì A là số nguyên
=>\(3n+9⋮n-4=>3n-12+21⋮n-4=>3.\left(n-4\right)+21⋮n-4\)
Mà \(\text{3 . (n - 4)}⋮n-4\)
=>\(21⋮n-4=>n-4\inƯ\left(21\right)=\left\{-21;-7;-3;-1;1;3;7;21\right\}\)
(Vì n là số nguyên => n - 4 là 1 số nguyên)
=>\(n\in\left\{-17;-3;1;3;5;9;11;25\right\}\)
Ta có bảng sau:
| n | -17 | -3 | 1 | 3 | 5 | 9 | 11 | 25 |
| 3n + 9 | -42 | 0 | 12 | 18 | 24 | 36 | 42 | 84 |
| n - 4 | -21 | -7 | -3 | -1 | 1 | 3 | 7 | 21 |
| \(A=\dfrac{3n+9}{n-4}\) | 2 | 0 | -4 | -18 | 24 | 12 | 6 | 4 |
Vậy.....
b)Vì B là số nguyên
=>\(2n-1⋮n+5=>2n+10-11⋮n+5=>2\left(n+5\right)-11⋮n+5\)
Mà \(\text{2 ( n + 5)}⋮n+5\)
=>\(11⋮n+5=>n+5\in\left\{-11;-1;1;11\right\}\)
(Vì n là số nguyên=> n + 5 là số nguyên)
=> \(n\in\left\{-16;-6;-4;6\right\}\)
Ta có bảng sau:
| n | -16 | -6 | -4 | 6 |
| 2 n - 1 | -33 | -13 | -9 | 11 |
| n + 5 | -11 | -1 | 1 | 11 |
| \(B=\dfrac{2n-1}{n+5}\) | 3 | 13 | -9 |
1 |
Vậy.......
\(\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{1}{3}\right|\ge0\\\left|x+\dfrac{1}{4}\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|+\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{1}{4}\right|\ge0\)
\(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+\dfrac{1}{2}+x+\dfrac{1}{3}+x+\dfrac{1}{4}=4x\)
\(\Leftrightarrow3x+1=4x\)
\(\Leftrightarrow x=1\left(tm\right)\)
Vậy ..