Tìm giá trị nhỏ nhất A= | x-2| |x-5|
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\(A=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\\ A_{min}=4\Leftrightarrow x=1\\ B=2\left(x^2-3x\right)=2\left(x^2-2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{9}{2}\\ B=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\\ B_{min}=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\\ C=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\\ C_{max}=7\Leftrightarrow x=2\)
a,\(A=x^2-2x+5=\left(x^2-2x+1\right)+4=\left(x-1\right)^2+4\ge4\)
Dấu "=" \(\Leftrightarrow x=-1\)
b,\(B=2\left(x^2-3x\right)=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)
Dấu "=" \(\Leftrightarrow x=\dfrac{3}{2}\)
c,\(=C=-\left(x^2-4x-3\right)=-\left[\left(x^2-4x+4\right)-7\right]=-\left(x-2\right)^2+7\le7\)
Dấu "=" \(\Leftrightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(A=\left|x-8\right|+\left|x+2\right|+\left|x+5\right|+\left|x+7\right|\ge\left|-x+8-x-2+x+5+x+7\right|=18\)
Dấu bằng xảy ra khi \(-5\le x\le-2\)
\(B=\left|x+3\right|+\left|x-5\right|+\left|x-2\right|\ge\left|x+3-x+5\right|+\left|x-2\right|=8+\left|x-2\right|\ge8\)
Dấu bằng xảy ra khi \(x=2\)
\(C=\left|x+5\right|-\left|x-2\right|\le\left|x+5+2-x\right|=7\)
Dấu bằng xảy ra khi \(x\ge2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(\left|x+2\right|\ge0\Rightarrow A=\left|x+2\right|+50\ge50\)
Dấu "=" xảy ra khi x=-2
Vậy GTNN của A=50 khi x=-2
b, Ta có: \(\left|x-100\right|\ge0;\left|y+200\right|\ge0\Rightarrow\left|x-100\right|+\left|y+200\right|\ge0\Rightarrow B=\left|x-100\right|+\left|y+200\right|-1\ge-1\)
Dấu "=" xảy ra khi x=100,y=-200
Vậy GTNN của B=-1 khi x=100,y=-200
c, Đặt C = 2015-|x+5|
Ta có: \(\left|x+5\right|\ge0\Rightarrow-\left|x+5\right|\le0\Rightarrow C=2015-\left|x+5\right|\le2015\)
Dấu "=" xảy ra khi x=-5
Vậy GTLN của C = 2015 khi x = -5
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2
Ta có:
\(A=\left|x-102\right|+\left|2-x\right|\Rightarrow A\ge\left|x-102+2-x\right|=-100\Rightarrow GTNNcủaAlà-100\)đạt được khi \(\left|x-102\right|.\left|2-x\right|=0\)
Trường hợp 1: \(x-102>0\Rightarrow x>102\)
\(2-x>0\Rightarrow x< 2\)
\(\Rightarrow102< x< 2\left(loại\right)\)
Trường hợp 2:\(x-102< 0\Rightarrow x< 102\)
\(2-x< 0\Rightarrow x>2\)
\(\Rightarrow2< x< 102\left(nhận\right)\)
Vậy GTNN của A là -100 đạt được khi 2<x<102.
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(A=3.\left|x+\frac{-2}{5}\right|+\frac{5}{2}\)
Ta có : \(\left|x+\frac{-2}{3}\right|\ge0\)
\(3.\left|x+\frac{-2}{3}\right|\ge0\)
\(3.\left|x+\frac{-2}{3}\right|+\frac{5}{2}\ge\frac{5}{2}\)
\(\Rightarrow Min_A=\frac{5}{2}\)
\(\Leftrightarrow3.\left|x+\frac{-2}{3}\right|=0\)
\(\Leftrightarrow\left|x+\frac{-2}{5}\right|=0\)
\(\Leftrightarrow x+\frac{-2}{5}=0\)
\(\Leftrightarrow x=\frac{2}{5}\)
`Answer:`
1.
Do \(\left|x-\frac{2}{5}\right|\ge0\forall x\)
\(\Rightarrow3.\left|x-\frac{2}{5}\right|\ge0\forall x\)
\(\Rightarrow3.\left|x-\frac{2}{5}\right|+\frac{5}{2}\ge\frac{5}{2}\forall x\)
Dấu "=" xảy ra khi \(\left|x-\frac{2}{5}\right|=0\Leftrightarrow x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}\)
Vậy \(3.\left|x-\frac{2}{5}\right|+\frac{5}{2}\) đạt giá trị nhỏ nhất \(=\frac{5}{2}\Leftrightarrow x=\frac{2}{5}\)
2.
Do \(\left|x-\frac{1}{2}\right|\ge0\forall x\)
\(\Rightarrow\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow A\ge\frac{3}{4}\)
Dấu "=" xảy ra khi \(\left|x-\frac{1}{2}\right|=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy giá trị nhỏ nhất của \(A=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(A=5-8x+x^2=-8x+x^2+6-11\)
\(=\left(x-4\right)^2-11\)
Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy Amin = - 11 <=> x = 4
\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)
\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy Bmax = 9 <=> x = - 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left|x-5\right|+\left|x+3\right|=\left|5-x\right|+\left|x+3\right|\)
\(\ge\left|5-x+x+3\right|=8\)
Dấu "=" xảy ra <=> \(\left(5-x\right)\left(x+3\right)\ge0\)
<=> \(-3\le x\le5\)
Vậy MIN \(A=8\) khi \(-3\le x\le5\)
Ta có :
A = | x - 2 | + | x - 5 |
A = | 2 - x | + | x - 5 | \(\ge\)| 2 - x + x - 5 | = 7
\(\Rightarrow\)Amin = 7 khi ( 2 - x ) . ( x - 5 ) \(\ge\)0 hay 2 \(\le\)x \(\le\)5