Áp dụng Bất đẳng thức trị tuyệt đối:
`a) |x - 2004| + |x - 2005|`
`= |x - 2004| + |2005 - x|`
`>= |x - 2004 + 2005 - x|`
`= |1| `
`= 1`
Vậy ...
`b) |x - 2| + |x-9| + 1945`
`= |x - 2| + |9 - x| + 1945`
`>= |x - 2 + 9 - x| + 1945`
`= |7| + 1945`
`= 7 + 1945`
`= 1952`
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`c) -|x-7|- |y+13| + 1945 `
Ta có: `{(|x-7| >=0),(|y+13| >=0):}`
`=> {(-|x-7| <=0),(-|y+13| <=0):}`
`=> -|x-7| -|y+13| <= 0`
`=> -|x-7|- |y+13| + 1945 <= 1945`
Dấu = có khi:
`{(x-7 =0),(y+13=0):}`
`<=> {(x=7),(y=-13):}`
Vậy ...
\(A=\left|x-2004\right|+\left|x-2005\right|=\left|x-2004\right|+\left|2005-x\right|\ge\left|x-2004+2005-x\right|=1\)
\(\Rightarrow A\left(min\right)=1\)
\(B=\left|x-2\right|+\left|x-9\right|+1945=\left|x-2\right|+\left|9-x\right|+1945\ge\left|x-2+9-x\right|+1945=1952\)
\(\Rightarrow B\left(min\right)=1952\)
