\(A=0,6+\left|\dfrac{1}{2}-x\right|\\ Vì:\left|\dfrac{1}{2}-x\right|\ge\forall0x\in R\\ Nên:A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\forall x\in R\\ Vậy:min_A=0,6\Leftrightarrow\left(\dfrac{1}{2}-x\right)=0\Leftrightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\\ Vì:\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\\ Nên:B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\forall x\in R\\ Vậy:max_B=\dfrac{2}{3}\Leftrightarrow\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow x=-\dfrac{1}{3}\)

= \(4x^2\)+\(20x\)+\(25\)+\(6x^2\)- \(8x\)- \(x^2\)-\(22\)

=\(9x^2\)+\(12x\)+\(3\)

=\(9x^2\)+\(12x\)+\(3\)

=\(9x^2\)+\(12x\)+\(4\)-\(1\)

=(\(3x\)+\(2\))²-\(1\)

vì (\(3x\)+\(2\))² >-0

=>.................-\(1\)>-(-1)

(>- là > hoặc =)

=> GTNN của M= -1 khi và chỉ khi \(3x\)+\(2\)=\(0\)

..................................

\(A=\left|x-3\right|+\left|y+3\right|+2016\)

\(\left|x-3\right|\ge0\)

\(\left|y+3\right|\ge0\)

\(\Rightarrow\left|x-3\right|+\left|y+3\right|+2016\ge2016\)

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

a ) \(A=0,6+\left|\dfrac{1}{2}-x\right|\)

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

\(\Leftrightarrow0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\)

Vậy GTNN là 0,6 khi \(x=\dfrac{1}{2}.\)

- Đề ghi ko hiểu ?

b ) \(\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)

Ta có : \(\left|2x+\dfrac{2}{3}\right|\ge0\)

\(\Leftrightarrow\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)

Vậy GTNN là \(\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)

\(A=0,6+\left|\dfrac{1}{2}-x\right|\)

\(\left|\dfrac{1}{2}-x\right|\ge0\forall x\in R\)

\(A=0,6+\left|\dfrac{1}{2}-x\right|\ge0,6\)

Dấu "=" xảy ra khi:

\(\left|\dfrac{1}{2}-x\right|=0\Rightarrow x=\dfrac{1}{2}\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\)

\(\left|2x+\dfrac{2}{3}\right|\ge0\forall x\in R\)

\(B=\dfrac{2}{3}-\left|2x+\dfrac{2}{3}\right|\le\dfrac{2}{3}\)

Dấu "=" xảy ra khi:

\(\left|2x+\dfrac{2}{3}\right|=0\Leftrightarrow2x=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{1}{3}\)

Gọi 1/4 số a là 0,25 . Ta có :

                   a . 3 - a . 0,25 = 147,07

                   a . (3 - 0,25) = 147,07 ( 1 số nhân 1 hiệu )

                      a . 2,75 = 147,07

                         a = 147,07 : 2,75

                          a = 53,48

mình nha

a) \(A=3\left|2x-\dfrac{3}{2}\right|+2021^0=3\left|2x-\dfrac{3}{2}\right|+1\ge1\)

\(minA=1\Leftrightarrow2x=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{4}\)

b) \(B=2\left|x-6\right|+3\left(2y-1\right)^2+2021^0=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\)

\(minB=1\Leftrightarrow\) \(\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=3\left|2x-\dfrac{3}{2}\right|+1\ge1\\ A_{min}=1\Leftrightarrow2x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{4}\\ B=2\left|x-6\right|+3\left(2y-1\right)^2+1\ge1\\ B_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=\dfrac{1}{2}\end{matrix}\right.\)

\(A=\left|2x-2\right|+\left|2x-2013\right|\)

\(A=\left|2x-2\right|+\left|2013-2x\right|\)

\(A\ge\left|2x-2+2013-2x\right|\)

\(A\ge2011\)Dấu "=" xảy ra khi: \(1\le x\le\frac{2013}{2}\)

A=|2x-2|+|2x-2013|

ta có |2x-2|=|2-2x|>hoặc=2-2x

. |2x-2013|>hoặc=2x-2013

=) A> hoặc = 2-2x+2x-2013

A> hoặc = -2011

A=|2x-2|+|2x-2013|

ta có |2x-2|=|2-2x|>hoặc=2-2x

. |2x-2013|>hoặc=2x-2013

=) A> hoặc = 2-2x+2x-2013

A> hoặc = -2011

 ..