giá trị tuyệt đối của x+5 + (3y-4)^2018=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)
Lại có : \(\left\{{}\begin{matrix}\left|x+5\right|\ge0\\\left(3y-4\right)^{2018}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left|x+5\right|+\left(3y-4\right)^{2018}\ge0\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left|x+5\right|=0\\\left(3y-4\right)^{2018}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-5=0\\3y-4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{4}{3}\end{matrix}\right.\)
Vậy ...
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\(\frac{\left(-2\right)^3}{5}.\left|\frac{1}{4}-1+2018^0\right|\)
\(=\frac{-8}{5}.\frac{1}{4}\)
\(=-\frac{2}{5}\)
\(\frac{\left(-2\right)^3}{5}\)x | \(\frac{1}{4}\)- 1| + 2018 mũ 0
![](https://rs.olm.vn/images/avt/0.png?1311)
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 |
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
| x | - | 2 | = 5
=> | x | - 2 = 5
=> | x \ = 7
=> \(\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
3 | x | = 18
=> | x | = 6
=> \(\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
2 | x | - 5 = 7
=> | x | = 7 + 5
=> | x | = 12
=> \(\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
| x | : 3 - 1 = | - 4 |
=> | x | : 3 - 1 = 4
=> | x | : 3 = 5
=> | x | = 15
=> \(\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
\(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)
Ta có \(\left|x+5\right|\ge0\forall x;\left(3y-4\right)^{2018}\ge0\forall y\)
Mà \(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\3y-4=0\end{cases}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{4}{3}\end{cases}}}\)
Vậy x = -5; y = 4/3