a) Ta có : \(A=\left|x+1\right|+\left|y-2\right|\)

\(\ge\left|x+1+y-2\right|\)

\(=\left|x+y-1\right|=\left|5-1\right|=\left|4\right|=4\)

Dấu "=" xảy ra <=> (x + 1)(y - 2) \(\ge\)0

Vậy Min A = 4 <=>  (x + 1)(y - 2) \(\ge\)0

\(A=\left|2x-1\right|+\left|2x-3\right|\)

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

\(\Rightarrow A\ge2\)

Dấu '' = '' xảy ra khi

\(\left(2x-1\right)\left(3-2x\right)\ge0\)

\(\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

Vậy Min A = 2 \(\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

a, \(A=\left|x+1\right|+\left|y-2\right|\)

\(A=\left|x+1\right|+\left|5-x-2\right|\)

\(A=\left|x+1\right|+\left|3-x\right|\ge x+1+3-x=4\)

Dấu " = " sảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\3-x\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le3\)

A = /2*-5-3/1+/2*-5

trả lời đê

mk sắp đi học dùi

\(0,5\left(\frac{3}{2}-2x\right)=\frac{1}{4}+\frac{2}{3}x\)

  \(\frac{3}{4}-x-\frac{1}{4}-\frac{2}{3}x=0\)

     \(\frac{1}{2}-\frac{5}{3}x=0\)

     \(\frac{5}{3}x=\frac{1}{2}\)

    \(x=\frac{3}{10}=0,3\)

\(\left|x-1\right|+2C=\left|x-1,5\right|+\left|1-x\right|\\ \Leftrightarrow\left|x-1\right|+2C=\left|x-1,5\right|+\left|x-1\right|\\ \Rightarrow2C=\left|x-1,5\right|\ge0\\ \Rightarrow C\ge0\)

Để C=0 thì

\(\left|x-1,5\right|=0\\ \Leftrightarrow x-1,5=0\\ \Leftrightarrow x=1,5\)

Vậy...

cái này sai r mk xóa nhé

Đề full ko phải vệ,có lẽ bạn đó viết quá gần