\(D=2015-5\left|x-386\right|-5\left|x-389\right|\)

\(D=2015-5\left(\left|x-386\right|+\left|389-x\right|\right)\)

\(D\le2015-5\left|x-386+389-x\right|\)

\(D\le2015-15=2000\)

Dấu "=" xảy ra khi: \(386\le x\le389\)

\(M=2016-\left|x-2015\right|-\left|x-1975\right|-\left|x-1945\right|\)

\(M=2016-\left(\left|x-2015\right|+\left|x-1975\right|+\left|x-1945\right|\right)\)

Đặt: \(L=\left|x-2015\right|+\left|x-1975\right|+\left|x-1945\right|\)

\(L=\left|x-2015\right|+\left|1945-x\right|+\left|x-1975\right|\)

\(L\ge\left|x-2015+1945-x\right|+\left|x-1975\right|\)

\(L\ge70+\left|x-1975\right|\ge70\)

Suy ra: \(M-L\le2016-70=1946\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}1945\le x\le2015\\x=1975\end{cases}}\Leftrightarrow x=1975\)

a. ta có \(\left|x-386\right|+\left|x-389\right|=\left|x-386\right|+\left|389-x\right|\ge\left|x-386+389-x\right|=3\)

\(\Rightarrow D\le2015-5\times3=2000\)

b. ta có \(\left|x-30\right|+\left|x-4\right|=\left|30-x\right|+\left|x-4\right|\ge\left|30-x+x-4\right|=26\)

\(\Rightarrow E\le\frac{51350}{26}=1975\)

Ta có : \(\left|x+2\right|+5\ge5\forall x\)

Nên : \(\frac{1}{\left|x+2\right|+5}\le\frac{1}{5}\)

<=> \(\frac{10}{\left|x+2\right|+5}\le\frac{10}{5}=2\)

Vậy A_max = 2 khi x = -2