Question

Tìm giá trị nhỏ nhất của 

  \(A=|x+2018|+|2019-x|\)

\(B=|2x+2018|+|2x-2019|\)

Answer 1

Tham khảo nha nhóc 

https://olm.vn/hoi-dap/detail/223396249611.html

Tương tự à 

Answer 2

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

\(\Rightarrow A=\left|x+2018\right|+\left|2019-x\right|\ge\left|\left(x+2018\right)+\left(2019-x\right)\right|=4037\)

\(\Rightarrow A_{min}=4037\)(Dấu "="\(\Leftrightarrow x\le2019\))

Answer 3

Kiệt sai rồi em

Đáp án Min A=4037 \(\Leftrightarrow-2018\le x\le2019\)

Answer 4

Ta có: \(B=\left|2x+2018\right|+\left|2x-2019\right|=\left|2x+2018\right|+\left|2019-2x\right|\)

Áp dụng BĐT \(\left|x\right|+\left|y\right|\ge\left|x+y\right|\)

\(B=\left|2x+2018\right|+\left|2019-2x\right|\ge\left|2x+2018+2019-2x\right|=4037\)

\(\Rightarrow B_{min}=4037\)

Answer 5

Ck Tiểu Ngư Cảm ơn anh nhé. E nhầm

Answer 6

\(A_{min}=4037\)(Dấu"="\(\Leftrightarrow-2018\le x\le2019\))

Answer 7

\(A=|x+2018|+|2019-x|\)\(\ge\)\(|x+2018+2019-x|\)\(=|4037|=4037\)

\(\Leftrightarrow\left(x+2018\right)\left(x+2019\right)\ge0\)

\(.........\)

\(B=|2x+2018|+|2x-2019|\)\(=|2x+2018|+|2019-2x|\)

\(\ge|2x+2018+2019-2x|\)\(=|4037|=4037\)

\(\Leftrightarrow\left(2x+2018\right)\left(2019-2x\right)\ge0\)

\(.......\)

Answer 8

Thank all nha 

Mặc dù bài này ko phải tui post