Question

Giá trị nhỏ nhất của biều thức A=  |x + y-2|+(2x-1)²⁰²²

 

Answer 1

Ta có: |x+y-2|≥0\(\forall\)x, y

Dấu "=" xảy ra \(\Leftrightarrow x+y-2=0\)

\(\left(2x-1\right)^{2022}\ge0\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(\Rightarrow A=\left|x+y-2\right|+\left(2x-1\right)^{2022}\ge0\)

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

Vậy \(A_{min}=0\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)