Bài 2 : 

a) \(A=3,7+\left|4,3-x\right|\ge3,7\)

Min A = 3,7 \(\Leftrightarrow x=4,3\)

b) \(B=\left|3x+8,4\right|-14\ge-14\)

Min B = -14 \(\Leftrightarrow x=\frac{-14}{5}\)

c) \(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Min C = 17,5 \(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{-3}{2}\end{cases}}\)

d) \(D=\left|x-2018\right|+\left|x-2017\right|\)

\(D=\left|2018-x\right|+\left|x-2017\right|\ge\left|2018-x+x-2017\right|=1\)

Min D =1 \(\Leftrightarrow\left(2018-x\right)\left(x-2017\right)\ge0\)

\(\Leftrightarrow2017\le x\le2018\)

\(A=3,7+\left|4,3-x\right|\)

Ta có \(\left|4,3-x\right|\ge0\Leftrightarrow A=3,7+\left|4,3-x\right|\ge3,7\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|4,3-x\right|=0\Leftrightarrow4,3-x=0\Leftrightarrow x=4,3\)

\(B=\left|3x+8,4\right|-14\)

Ta có \(\left|3x+8,4\right|\ge0\Leftrightarrow B=\left|3x+8,4\right|-14\ge-14\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left|3x+8,4\right|=0\Leftrightarrow3x=-8,4\Leftrightarrow x=2,8\)

\(C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Ta có \(\hept{\begin{cases}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{cases}}\Leftrightarrow C=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu '' = '' xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

\(D=\left|x-2018\right|+\left|x-2017\right|\)

\(\Leftrightarrow D=\left|x-2018\right|+\left|2017-x\right|\)

Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)ta có

\(D\ge\left|x-2018+2017-x\right|=\left|-1\right|=1\)

Dấu '' = '' xảy ra \(\Leftrightarrow\left(2017-x\right)\left(x-2018\right)\ge0\Leftrightarrow2018\ge x\ge2017\)

0.84

a)A=3+|1-x|

       Vì \(\left|1+x\right|\ge0\)

            Suy ra:\(3+\left|1+x\right|\ge3\)

Dấu = xảy ra khi 1+x=0;x=-1

                   Vậy Min A=3 khi x=-1

b)B=|4,3-x|+3,7

        Vì \(\left|4,3-x\right|\ge0\)

                Suy ra:\(\left|4,3-x\right|+3,7\ge3,7\)

Dấu = xảy ra khi 4,3-x=0;x=4,3

            Vậy Min B=3,7 khi x=4,3
 

C=2|3x+8,4|-14,2

     Vì \(2\left|3x+8,4\right|\ge0\)

           Suy ra:\(2\left|3x+8,4\right|-14,2\ge-14,2\)

                      Dấu = xảy ra khi 3x+8,4=0;x=-2,8

Vậy Min C=-14,2 khi x=-2,8

d)D=|x+1|+2|3x+8,4|-14,2

         Vì \(\left|x+1\right|\ge0\)

               \(2\left|3x+8,4\right|\ge0\)

           Suy ra:\(\left|x+1\right|+2\left|3x+8,4\right|-14,2\ge-14,2\)

                      Dấu = xảy ra khi x+1=0;x=-1

                                                3x+8,4=0;x=-2,8

Vậy Min D=-14,2 khi x=-2,8;-1

\(A=605x8,4=5082\)

     2. Thay A = 3509, có:

\(y=3509:605=\frac{29}{5}=5,8\)

\(\Rightarrow X< 15:4,5\)

\(\Rightarrow X< 3\frac{1}{3}\)

\(\Rightarrow X=0;1;2;3\)

A= 605 × 8,4 = 5082

y = 3509 ÷ 605 = 3,8

X = 0, x = 1, x = 2 hoặc x = 3

có 

A=|3x+8,4|-14,2

=>A=|3x+8,4|-14,2≥-14,2

dấu "=" xảy ra khi |3x+8,4|=0 =>x=2,8

vs minA=-14,2 khi x=2,8

mình không pk đúng hay sai nx 

a) \(A=\left|3x+8,4\right|-14,2\ge-14,2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow3x+8,4=0\Leftrightarrow x=-2,8\)

Vậy........

b) \(B=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

\(B\ge17,5\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}}\)

Vậy.........