Question

[Bui Trinh Minh Ngoc]

cho |2x-1|+(3y+2)² be hon hoac bang 0

tinh S=x²+y²-xy

Answer 1

[kudo shinichi]

Ta có: \(\hept{\begin{cases}\left|2x-1\right|\ge0\forall x\\\left(3y+2\right)^2\ge0\forall y\end{cases}\Rightarrow\left|2x-1\right|+\left(3y+2\right)^2\ge0\forall x;y}\)

Mà \(\left|2x-1\right|+\left(3y+2\right)^2\le0\)

Dấu = xảy ra \(\Rightarrow\hept{\begin{cases}\left|2x-1\right|=0\\\left(3y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\3y+2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-2}{3}\end{cases}}}\)

\(\Rightarrow S=x^2+y^2-xy=\left(\frac{1}{2}\right)^2+\left(\frac{-2}{3}\right)^2-\left(\frac{1}{2}.\frac{-2}{3}\right)\)

\(S=\frac{1}{4}+\frac{4}{9}+\frac{1}{3}\)

\(S=\frac{9}{36}+\frac{16}{36}+\frac{12}{36}\)

\(S=\frac{37}{36}\)

Answer 2

[Phùng Minh Quân]

Ta có : 

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

\(\left(3y+2\right)^2\ge0\)

\(\Rightarrow\)\(\left|2x-1\right|+\left(3y+2\right)^2\ge0\)

Mà \(\left|2x-1\right|+\left(3y+2\right)^2\le0\) ( Giả thiết ) 

Do đó : \(\left|2x-1\right|+\left(3y+2\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}\left|2x-1\right|=0\\\left(3y+2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1=0\\3y+2=0\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}2x=1\\3y=-2\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{-2}{3}\end{cases}}}\)

Thay \(x=\frac{1}{2}\) và \(y=\frac{-2}{3}\) vào \(S=x^2+y^2-xy\) ta được : 

\(S=\left(\frac{1}{2}\right)^2+\left(\frac{-2}{3}\right)^2-\frac{1}{2}.\frac{-2}{3}\)

\(S=\frac{1}{4}+\frac{4}{9}+\frac{1}{3}\)

\(S=\frac{3}{4}\)

Vậy \(S=\frac{3}{4}\)

Chúc bạn học tốt ~