\(\left|x+5\right|+\left(3y-4\right)^{2018}=0\)

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

\(\frac{\left(-2\right)^3}{5}.\left|\frac{1}{4}-1+2018^0\right|\)

\(=\frac{-8}{5}.\frac{1}{4}\)

\(=-\frac{2}{5}\)

\(\frac{\left(-2\right)^3}{5}\)x | \(\frac{1}{4}\)- 1| + 2018 mũ 0

\(\frac{\left(-2\right)^3}{5}.\left|\frac{1}{4}-1\right|+2018^0\)

\(=-\frac{8}{5}.\frac{-3}{4}+1\)

\(=\frac{6}{5}+1\)

\(=\frac{11}{5}\)

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) \(x-2xy+y-3=0\)

\(\Leftrightarrow x\left(1-2y\right)+y-\frac{1}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow2x\left(\frac{1}{2}-y\right)-\left(\frac{1}{2}-y\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(\frac{1}{2}-y\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=5\)

| 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

=> \(\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)