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\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
a) \(P=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)
*TH1: \(x< 2016\):
\(P=2016-x+2017-x+2018-x=6051-3x>6051-3\cdot2016=3\)
*TH2: \(2016\le x< 2017\):
\(P=x-2016+2017-x+2018-x=2019-x>2019-2017=2\)
*TH3: \(2017\le x< 2018\):
\(P=x-2016+x-2017+2018-x=x-2015\ge2017-2015=2\)(Dấu "=" xảy ra khi x = 2017)
*TH4: \(x\ge2018\):
\(P=x-2016+x-2017+x-2018=3x-6051\ge3\cdot2018-6051=3\)(Dấu "=" xảy ra khi x = 2018)
Vậy GTNN của P là 2 khi x = 2017.
b) \(x-2xy+y-3=0\)
\(\Leftrightarrow x\left(1-2y\right)+y-\frac{1}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow2x\left(\frac{1}{2}-y\right)-\left(\frac{1}{2}-y\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{1}{2}-y\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=5\)
| 2x-1 | 5 | -5 | 1 | -1 |
| 1-2y | 1 | -1 | 5 | -5 |
| x | 3 | -2 | 1 | 0 |
| y | 0 | 1 | -2 | 3 |
\(\left|x-y\right|+\left|y+\frac{9}{25}\right|=0\)
Ta có : \(\hept{\begin{cases}\left|x-y\right|\ge0\\\left|y+\frac{9}{25}\right|\ge0\end{cases}}\Rightarrow\left|x-y\right|+\left|y+\frac{9}{25}\right|\ge0\forall x,y\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-y=0\\y+\frac{9}{25}=0\end{cases}}\Rightarrow\hept{\begin{cases}x-y=0\\y=-\frac{9}{25}\end{cases}\Rightarrow}x=y=-\frac{9}{25}\)
Vậy giá trị của biểu thức = 0 khi x = y = -9/25
Ta có: /x-y/ \(\ge\) 0 với mọi x,y
/y+9/36/ \(\ge\) 0 với mọi y
=> /x-y/ + /y+9/36/ \(\ge\) 0 vs mọi x,y
Ta có: /x-y/ + /y+9/36/ \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|y+\dfrac{9}{36}\right|\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{36}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=-\dfrac{9}{36}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{36}\\y=-\dfrac{9}{36}\end{matrix}\right.\)
Vậy x = -9/36 và y = -9/36