Question

[D.Khánh Đỗ]

Tìm GTNN của biểu thức:

A= 4x² + 3y² - 6xy + 6x - 12y +20

Answer 1

[❥一Ǥoɗɗεşş øf iηէεℓℓεcէ︵♛╾━╤デ╦︻]

\(A=4x^2-6x\left(x-y\right)+3y^2-12y+20\)

\(A=\left(2x\right)^2-2.2x.\frac{3}{2}y+\left(\frac{3}{2}y\right)^2-\frac{9}{4}y^2+3y^2-12y+20\)

\(A=\left(2x-\frac{3}{4}y\right)^2+\frac{3}{4}y^2-12y+432-432+20\)

\(A=\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{4}y^2-2.\frac{1}{2}.12+12^2\right)-432+20\)

\(\Rightarrow A=\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{2}y-12\right)^2-412\)

Ta có:\(\hept{\begin{cases}\left(2x-\frac{3}{4}y\right)^2\ge0\\\left(\frac{1}{2}y-12\right)^2\ge0\Rightarrow3\left(\frac{1}{2}y-12\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{2}y-12\right)^2-412\ge-412\)

\(\Rightarrow A_{min}=-412\)đạt được khi

i\(\hept{\begin{cases}\left(2x-\frac{3}{4}y\right)^2=0\\\left(\frac{1}{2}y-12\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-\frac{3}{4}y=0\\\frac{1}{2}y-12=0\end{cases}\Leftrightarrow}\hept{\begin{cases}2x=\frac{3}{4}y\\\frac{1}{2}y=12\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\y=24\end{cases}}}\)