Question

Tìm GTNN của

\(A=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Answer 1

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Answer 2

\(A=|4x-3|+|5y+7,5|+17,5\)

\(|4x-3|\ge0\)

\(|5y+7,5|\ge0\)

\(\Leftrightarrow|4x-3|+|5y+7,5|+17,5\ge17,5\)

Vậy \(MaxA=17,5\)khi \(\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

Answer 3

Ta có \(\left|4x-3\right|\ge0\)\(\forall x\)

          \(\left|5y+7,5\right|\ge0\forall x\)

=> \(\left|4x-3\right|+\left|5y+7,5\right|\ge0\forall x,y\)

=>\(\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\forall x,y\)

hay \(A\ge17,5\)

\(MinA=17,5\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=-\frac{3}{2}\end{cases}}\)

Answer 4

Bạn Kaito Kid sai rùi, GTNN phải là Min chứ!

Answer 5

Ta có : \(\hept{\begin{cases}\left|4x-3\right|\ge0\\\left|5y+7,5\right|\ge0\end{cases}\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|\ge0}\)

\(\Leftrightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

Dấu '' = '' xảy ra 

\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\y=\frac{-3}{2}\end{cases}}}\)

Vậy.................................