Bài 2: 

a) Ta có: \(\left|2x-5\right|\ge0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

\(\Leftrightarrow-\left|2x-5\right|+3\le3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

\(a,\left|3x-1\right|=\left|5-2x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}3x-1=5-2x\\3x-1=2x-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=6\\x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{6}{5}\\x=-4\end{cases}}\)

b,\(\left|2x-1\right|+x=2\)

\(\Leftrightarrow\left|2x-1\right|=2-x\)

Điều kiện \(2-x\ge0\Leftrightarrow x\le2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=2-x\\2x-1=x-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=3\\x=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\left(\text{nhận}\right)\\x=-1\left(\text{nhận}\right)\end{cases}}}\)

c.\(A=0,75-\left|x-3,2\right|\)

Vì \(\left|x-3,2\right|\ge0\Rightarrow0,75-\left|x-3,2\right|\le0,75\)

Dấu "=' xảy ra \(\Leftrightarrow x-3,2=0\Leftrightarrow x=3,2\)

Vậy Max A = 0,75 khi x = 3,2

\(d,B=2.\left|x+1,5\right|-3,2\)

Vì 2. |x + 1,5| ≥ 0 => B ≥ -3,2

Dấu " = ' xảy ra khi \(2\left|x+1,5\right|=0\)

\(\Leftrightarrow x+1,5=0\Leftrightarrow x=-1,5\)

Vậy Min B = -3,2 khi x = -1,5

a) \(A=\left|x-5\right|+\left|x-7\right|=\left|x-5\right|+\left|7-x\right|\ge\left|x-5+7-x\right|=\left|2\right|=2\)

\(minA=2\Leftrightarrow\)\(7\ge x\ge5\)

b) \(B=\left|2x+1\right|+\left|2x-2\right|=\left|2x+1\right|+\left|2-2x\right|\ge\left|2x+1+2-2x\right|=\left|3\right|=3\)

\(minB=3\Leftrightarrow1\ge x\ge-\dfrac{1}{2}\)

a) Do \(\left|1+2x\right|\ge0\Rightarrow\dfrac{-1}{4}\left|1+2x\right|\le0\)

\(\Rightarrow A=2,25-\dfrac{1}{4}\left|1+2x\right|\le2,25\)

\(maxA=2,25\Leftrightarrow x=-\dfrac{1}{2}\)

b) Do \(\left|2x-3\right|\ge0\Rightarrow3+\dfrac{1}{2}\left|2x-3\right|\ge3\)

\(\Rightarrow B=\dfrac{1}{3+\dfrac{1}{2}\left|2x-3\right|}\le\dfrac{1}{3}\)

\(maxB=\dfrac{1}{3}\Leftrightarrow x=\dfrac{3}{2}\)

mình ghi nhầm đề bài là Tìm giá trị lớn nhất nhé

\(A=-\left|x-7\right|+2\le2\\ A_{max}=2\Leftrightarrow x-7=0\Leftrightarrow x=7\\ B=-5-\left|2x+3\right|\le-5\\ A_{max}=-5\Leftrightarrow2x+3=0\Leftrightarrow x=-\dfrac{3}{2}\)

a: \(A=4x^2-4x+1-4=\left(2x-1\right)^2-4>=-4\forall x\)

Dấu '=' xảy ra khi x=1/2

a

C= |x-1| + |x-5|

Do x-1 + x-5 luôn > 0

=> x-1 + x-5 = 0

=> 2x -6 = 0

=> 2x = 6

=> x = 3

mình ghi nhầm, lớn hơn hoặc bằng 0 nha

a) Ta có C = |x - 1| + |x - 5| = |x - 1| + |5 - x| \(\ge\left|x-1+5-x\right|=\left|4\right|=4\)

=> Min C = 4

Dấu "=" xảy ra <=> (x -1)(5 - x) \(\ge0\)

Xét 2 trường hợp

TH1 : \(\hept{\begin{cases}x-1\ge0\\5-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge1\\x\le5\end{cases}}\Leftrightarrow1\le x\le5\)

TH2 : \(\hept{\begin{cases}x-1\le0\\5-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge5\end{cases}\Leftrightarrow x\in\varnothing}\)

Vậy Min C = 4 <=> \(1\le x\le5\)

2) a) Ta có \(\left|2x-5\right|\ge0\forall x\Leftrightarrow-\left|2x-5\right|\le0\forall x\)

=> \(C=3-\left|2x-5\right|\le3\forall x\)

=> Max C = 3

Dấu "=" xảy ra <=> 2x - 5 = 0 <=> x = 2,5

Vậy Max C = 3 <=> x = 2,5

b) Ta có \(2\left|x-1\right|\ge0\forall x\Leftrightarrow2\left|x-1\right|+3\ge3\forall x\)

=> D = \(\frac{1}{2\left|x-1\right|+3}\le\frac{1}{3}\forall x\)

=> Max D = 1/3 

Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1

Vậy Max D = 1/3 <=> x = 1

a = |2x-1/3|-7/4

   Do |2x-1/3| \(\ge\) 0

         |2x-1/3|-7/4 \(\ge\)  7/4 

Dấu = xảy ra <=> 2x-1/3=0. =>. x= 1/6

b    1/3|x-2|+2|3-1/2 y|+4

 Do |x-2| \(\ge\) 0

      |3-1/2y| \(\ge\) 0

   => 1/3|x-2|+2|3-1/2 y|+4 \(\ge\) 4

Dấu = xảy ra <=>\(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

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 \(x=\dfrac{1}{6}\)

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

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

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

\(\Leftrightarrow\left|x-2\right|\cdot\dfrac{1}{3}+\left|3-\dfrac{1}{2}y\right|\cdot2+4\ge4\forall x,y\)

Dấu '=' xảy ra khi x=2 và y=6