a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)
Ta có: \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|\ge0\forall x\)
\(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\)
Dấu = xảy ra khi :
\(\frac{x}{5}+\frac{23}{2}=0\Leftrightarrow\frac{x}{5}=-\frac{23}{2}\Leftrightarrow x=-\frac{115}{2}\)
\(y-\frac{14}{3}=0\Leftrightarrow y=\frac{14}{3}\)
Vậy ..............
Ta có:
a) \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)
\(\left|y-\frac{14}{3}\right|\ge0\forall y\)
=> \(\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{5}+\frac{23}{2}=0\\y-\frac{14}{3}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
Vậy Min của A = 2019 tại \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)
câu b tượng tự
\(b,B=\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{30}-11\)
Ta có : \(\left[x-\frac{5}{4}\right]^{20}\ge0\forall x\)
\(\left[y-\frac{4}{3}\right]^{30}\ge0\forall y\)
\(\Leftrightarrow\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{20}-11\ge-11\forall x,y\)
Dấu " = " xảy ra : \(\hept{\begin{cases}\left[x-\frac{5}{4}\right]^{20}=0\\\left[y-\frac{4}{3}\right]^{20}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\frac{5}{4}=0\\y-\frac{4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=\frac{4}{3}\end{cases}}\)
Vậy : ...
