A = |x| + 7

|x| >/ 0

=> A >/ 7

Vậy GTNN của A = 7 kh |x| = 0 <=> x= 0 

x thuoc tap hop rong vi:

ko co gia tri nao thoa man x

Ta có: 

15-|x-2|=12

     |x-2|=15-12

     |x-2|=3

=> x-2 \(\in\){-3;3}

Với x-2= -3

       x   = -3+2 

      x    = -1

Với x-2 = 3 

       x    = 3+2 

      x     = 5

Vậy x\(\in\){-1;5} thì 15-|x-2|=12

=> /x-2/=15-12

=> /x-2/=3

=> x-2={-3;3}

Nếu x-2=3 thì x=5

Nếu x-2=-3 thì x= -1

ta có:

\(\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|x-4\right|\)

\(=\left|x-1\right|+\left|x-2\right|+\left|y-3\right|+\left|4-x\right|\)

\(\ge\left|x-1+4-x\right|+\left|x-2\right|+\left|y-3\right|\)

\(=3+\left|x-2\right|+\left|y-3\right|\)

\(\ge3\)

Dấu "=" xả ra khi \(\hept{\begin{cases}\left(x-1\right)\left(4-x\right)\ge0\\\left|x-2\right|=0\\\left|y-3\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}1\le x\le4\cdot\\x=2\left(TM\cdot\right)\\y=3\end{cases}}\)

Vậy \(x=2;y=3\)

(x-1) + (x-2) + (x-3) + (x-4) = 3

(x+x+x+x) - (1+2+3+4) = 3

X x 4 - 10 = 3

X x 4 = 3 + 10

X x 4 = 13

x = 13 : 4

x = \(\frac{13}{4}\)

Ta có:\(\left|x+1\right|\ge0;\left|x-2\right|\ge0;\left|x+7\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x-2\right|+\left|x+7\right|\ge0\)

\(\Rightarrow5x-10\ge0\)

\(\Rightarrow5x\ge10\)

\(\Rightarrow x\ge2\)

\(\Rightarrow\left|x+1\right|=x+1\)

  \(\left|x-2\right|=x-2\)

  \(\left|x+7\right|=x+7\)

Ta có:\(\left|x+1\right|+\left|x-2\right|+\left|x+7\right|=5x-10\)

\(\Rightarrow x+1+x-2+x+7=5x-10\)

\(\Rightarrow\)\(3x+6=5x-10\)

\(\Rightarrow6+10=5x-3x\)

\(\Rightarrow2x=16\)

\(\Rightarrow x=8\)

Vậy x=8 thỏa mãn