a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)

hay \(x=\dfrac{1}{6}\)

Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)

b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)

\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)

Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)

\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

Vậy: \(B_{min}=4\) khi x=2 và y=6

a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)

b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)

c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)

d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)

có 

A=|3x+8,4|-14,2

=>A=|3x+8,4|-14,2≥-14,2

dấu "=" xảy ra khi |3x+8,4|=0 =>x=2,8

vs minA=-14,2 khi x=2,8

mình không pk đúng hay sai nx 

a) \(A=\left|3x+8,4\right|-14,2\ge-14,2\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow3x+8,4=0\Leftrightarrow x=-2,8\)

Vậy........

b) \(B=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

\(B\ge17,5\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}}\)

Vậy.........

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

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

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) Vì \(\left|4,3-x\right|\ge0\Rightarrow A=3,7+\left|4,3-x\right|\ge3,7\)

Dấu "=" xảy ra <=> \(\left|4,3-x\right|=0\Leftrightarrow4,3-x=0\Leftrightarrow x=4,3\)

Vậy A_min = 3,7 khi và chỉ khi x = 4,3

b) Vì \(\left|3x+8,4\right|\ge0\Rightarrow B=\left|3x+8,4\right|-14\ge-14\)

Dấu "=" xảy ra <=> \(\left|3x+8,4\right|=0\Leftrightarrow3x=-8,4\Leftrightarrow x=-2,8\)

Vậy B_Min = -14 khi và chỉ khi x = -2,8

c) Vì \(\left|4x-3\right|\ge0;\left|5y+7,5\right|\ge0\Rightarrow B=\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

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

Vậy C_Min = 17,5 khi và chỉ khi x = 3/4 và y = -1,5

d) D = |x-2018| + |x-2017| = |x-2018| + |2017-x| lớn hơn hoặc bằng |x-2018+2017-x| = |-1|=1

Dấu "=" xảy ra khi và chỉ khi (x-2018).(2017-x) lớn hơn hoặc bằng 0

              (Tự giải ra)

Vậy D_Min = 1 khi và chỉ khi ...