k) Vì \(\left|4x-3\right|\ge0\left(\forall x\right);\left|5y+7,5\right|\ge0\left(\forall y\right)\)

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{-3}{2}\end{cases}}}\)

Vậy C_Min = 17,5 khi và chỉ khi x = 3/4 và y = -3/2

n) Ta có: 

\(M=\left|x-2002\right|+\left|x-2001\right|=\left|x-2002\right|+\left|2001-x\right|\ge\left|x-2002+2001-x\right|=1\)

Dấu "=" xảy ra khi \(\left(x-2002\right)\left(2001-x\right)\ge0\) 

<=> x lớn hơn hoặc bằng 2002

Hoặc x bé hơn hoặc bằng 2001

Vậy M_Min =1

a)

• Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

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

\(\Rightarrow B\ge3\)

Dấu = khi \(\left(x-1\right)\left(x-4\right)\ge0\)\(\Rightarrow1\le x\le4\)

Vậy MinB=3 khi \(1\le x\le4\)

• Áp dụng tiếp Bđt kia ta có:

\(\left|1993-x\right|+\left|1994-x\right|\ge\left|1993-x+x-1994\right|=1\)

\(\Rightarrow C\ge1\)

Dấu = khi \(\left(x-1993\right)\left(x-1994\right)\ge0\)\(\Rightarrow1993\le x\le1994\)

Vậy MinC=1 khi \(1993\le x\le1994\)

• Ta thấy: \(\begin{cases}x^2\\\left|y-2\right|\end{cases}\ge0\)

\(\Rightarrow x^2+\left|y-2\right|\ge0\)

\(\Rightarrow x^2+\left|y-2\right|-5\ge-5\)

\(\Rightarrow D\ge-5\)

Dấu = khi \(\begin{cases}x=0\\y=2\end{cases}\)

Vậy MinD=-5 khi \(\begin{cases}x=0\\y=2\end{cases}\)

b)Ta thấy:

\(\begin{cases}\left|4x-3\right|\\\left| 5y+7,5\right|\end{cases}\ge0\)

\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|\ge0\)

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

\(\Rightarrow C\ge17,5\)

Dấu = khi \(\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}\)

Vậy MinC=17,5 khi \(\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}\)

c)Áp dụng Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x-2002\right|+\left|x-2001\right|\ge\left|x-2002+2001-x\right|=1\)

\(\Rightarrow M\ge1\)

Dấu = khi \(\left(x-2002\right)\left(x-2001\right)\ge0\)\(\Rightarrow2001\le x\le2002\)

Vậy MinM=1 khi \(2001\le x\le2002\)

Thanks

ta có \(\left|4x-3\right|\ge0;\left|5y+7,5\right|\ge0\Rightarrow E\ge17,5\)

dấu = xảy ra <=> \(\hept{\begin{cases}\left|4x-3\right|=0\\\left|5y+7,5\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0,75\\y=1,5\end{cases}}}\)

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 \(\Leftrightarrow\left(2017-x\right)\left(x-2018\right)\ge0\Leftrightarrow2018\ge x\ge2017\)

Vì : |4x-3| >= 0

       |5y+7,5| >= 0

nên |4x-3|+|5y+7,5|+17,5>= 0+0+17,5

hay E>= 17,5

Dấu " =" xảy ra khi: \(\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\)   <=>\(\hept{\begin{cases}4x=3\\5y=-7,5\end{cases}}\)

                          <=> \(\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

Vậy Min_E = 17,5 khi (x,y) = (\(\frac{3}{4}\); -1,5)