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\(A=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)
Ta thấy \(\left|4x-3\right|\ge0;\left|5y+7,5\right|\ge0\)
\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)
\(\Rightarrow A\ge17,5\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)
...
\(B=\left|x-2\right|+\left|x-6\right|+2017\)
\(=\left|x-2\right|+\left|6-x\right|+2017\)
Ta thấy \(\left|x-2\right|+\left|6-x\right|\ge\left|x-2+6-x\right|=4\)
\(\Rightarrow B\ge4+2017=2021\)
Dấu "=" xảy ra khi \(2\le x\le6\)
....
\(C=\left(2x+1\right)^{2020}-2019\)
Ta thấy \(\left(2x+1\right)^{2020}\ge0\)
\(\Rightarrow C=\left(2x+1\right)^{2020}-2019\ge-2019\)
Dấu "=" xảy ra khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)
....
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 \(2001\le x\le2002\)
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 \(\Leftrightarrow\left(2017-x\right)\left(x-2018\right)\ge0\Leftrightarrow2018\ge x\ge2017\)
Ta có: 17,5 - 2,3x = 7,5 - 14,3
=> 17,5 - 7,5 = -14,3x + 2,3x
=> 17,5 - 7.5 = x(-14,3 - 2,3)
=> 12x = 10
=> x = 10 : 12
=> x = 5/6