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\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)
\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)
1/ \(A=3\left|2x-1\right|-5\)
Ta có: \(\left|2x-1\right|\ge0\)
\(\Rightarrow3\left|2x-1\right|\ge0\)
\(\Rightarrow3\left|2x-1\right|-5\ge-5\)
Để A nhỏ nhất thì \(3\left|2x-1\right|-5\)nhỏ nhất
Vậy \(Min_A=-5\)
2.
a/\(A=5-I2x-1I\)
Ta thấy: \(I2x-1I\ge0,\forall x\)
nên\(5-I2x-1I\le5\)
\(A=5\)
\(\Leftrightarrow5-I2x-1I=5\)
\(\Leftrightarrow I2x-1I=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy GTLN của \(A=5\Leftrightarrow x=\frac{1}{2}\)
b/\(B=\frac{1}{Ix-2I+3}\)
Ta thấy : \(Ix-2I\ge0,\forall x\)
nên \(Ix-2I+3\ge3,\forall x\)
\(\Rightarrow B=\frac{1}{Ix-2I+3}\le\frac{1}{3}\)
\(B=\frac{1}{3}\)
\(\Leftrightarrow B=\frac{1}{Ix-2I+3}=\frac{1}{3}\)
\(\Leftrightarrow Ix-2I+3=3\)
\(\Leftrightarrow Ix-2I=0\)
\(\Leftrightarrow x=2\)
Vậy GTLN của\(A=\frac{1}{3}\Leftrightarrow x=2\)
\(A=\left|x-3\right|+\left|y+3\right|+2016\)
\(\left|x-3\right|\ge0\)
\(\left|y+3\right|\ge0\)
\(\Rightarrow\left|x-3\right|+\left|y+3\right|+2016\ge2016\)
Dấu ''='' xảy ra khi \(x-3=y+3=0\)
\(x=3;y=-3\)
\(MinA=2016\Leftrightarrow x=3;y=-3\)
\(\left(x-10\right)+\left(2x-6\right)=8\)
\(x-10+2x-6=8\)
\(3x=8+10+6\)
\(3x=24\)
\(x=\frac{24}{3}\)
x = 8
Ta có: |2x - 5| \(\ge\)0 \(\forall\)x
=> |2x - 5| + 1,(3) \(\ge\)1,(3)
hay |2x - 5| + 4/3 \(\ge\)4/3
Dấu "=" xảy ra <=> 2x - 5 = 0 <=> x = 5/2
Vậy Min F = 4/3 <=> x = 5/2
Ta có: G = |x - 3| + |x + 3/2|
G = |3 - x| + |x + 3/2| \(\ge\)|3 - x + x + 3/2| = |3/2| = 3/2
Dấu "=" xảy ra <=> (3 - x)(x + 3/2) \(\ge\)0
<=> -3/2 \(\le\)x \(\le\)3
Vậy MinG = 3/2 <=> -3/2 \(\le\)x \(\le\)3
Làm lại cho Edogawa Conan
\(G=\left|x-3\right|+\left|x+\frac{3}{2}\right|\)
\(G=\left|3-x\right|+\left|x+\frac{3}{2}\right|\ge\left|\left(3-x\right)+\left(x+\frac{3}{2}\right)\right|\)
\(=\frac{9}{2}\)
Vậy \(G_{min}=\frac{9}{2}\Leftrightarrow\left(3-x\right)\left(x+\frac{3}{2}\right)\ge0\)
\(Th1:\hept{\begin{cases}3-x\ge0\\x+\frac{3}{2}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge\frac{3}{2}\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le2\)
\(Th2:\hept{\begin{cases}3-x\le0\\x+\frac{3}{2}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le\frac{3}{2}\end{cases}}\left(L\right)\)
\(C=\left|2x+1\right|+\left|-2y-1\right|\ge\left|2x+1-2y-1\right|=2\left|x-y\right|=4\)
\(C_{min}=4\)