a)\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\)
Vì \(-\left|x+\frac{3}{2}\right|\)\(\le\)0
Suy ra:\(\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)
Dấu = xảy ra khi \(x+\frac{3}{2}=0\)
\(x=-\frac{3}{2}\)
Vậy Max A=\(\frac{1}{4}\) khi \(x=-\frac{3}{2}\)
b)\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)
Vì \(-\left|x-\frac{4}{3}\right|\le0;-\left|y+\frac{1}{2}\right|\le0\)
Suy ra:\(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)
Dấu = xảy ra khi \(x-\frac{4}{3}=0;x=\frac{4}{3}\)
\(y+\frac{1}{2}=0;y=-\frac{1}{2}\)
Vậy Max B=\(\frac{5}{3}\) khi \(x=\frac{4}{3};y=-\frac{1}{2}\)
a/ Ta có ; \(\left|x+\frac{3}{2}\right|\ge0\Rightarrow-\left|x+\frac{3}{2}\right|\le0\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)
Vậy BT đạt giá trị lớn nhất bằng 1/4 khi x = -3/2
b/ \(\begin{cases}\left|x-\frac{4}{3}\right|\ge0\\\left|y+\frac{1}{2}\right|\ge0\end{cases}\) \(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\le0\\-\left|y+\frac{1}{2}\right|\le0\end{cases}\)
\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)
\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)
Vậy BT đạt giá trị lớn nhất bằng 5/3 khi x = 4/3 , y = -1/2
a)Đặt \(A=\frac{1}{4}-\left|x+\frac{3}{2}\right|\)
Ta thấy: \(\left|x+\frac{3}{2}\right|\ge0\)
\(\Rightarrow-\left|x+\frac{3}{2}\right|\le0\)
\(\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}-0=\frac{1}{4}\)
\(\Rightarrow A\le\frac{1}{4}\)
Dấu = khi \(x=-\frac{3}{2}\)
Vậy MaxA=\(\frac{1}{4}\Leftrightarrow x=-\frac{3}{2}\)
b)Đặt \(B=\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)
Ta thấy: \(\begin{cases}\left|x-\frac{4}{3}\right|\\\left|y+\frac{1}{2}\right|\end{cases}\ge0\)
\(\Rightarrow\begin{cases}-\left|x-\frac{4}{3}\right|\\-\left|y+\frac{1}{2}\right|\end{cases}\)\(\le0\)
\(\Rightarrow-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le0\)
\(\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}-0=\frac{5}{3}\)
\(\Rightarrow B\le\frac{5}{3}\)
Dấu = khi \(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)
Vậy MaxB=\(\frac{5}{3}\Leftrightarrow\)\(\begin{cases}x=\frac{4}{3}\\y=-\frac{1}{2}\end{cases}\)
a) \(\frac{1}{4}-\left|x+\frac{3}{2}\right|\) có GTLN
=> \(\left|x+\frac{3}{2}\right|\) có GTNN
=> \(\left|x+\frac{3}{2}\right|=0\)
=> \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)
a) \(\frac{1}{4}-\left|x+\frac{3}{2}\right|\)
Có: \(\left|x+\frac{3}{2}\right|\ge0\Rightarrow\frac{1}{4}-\left|x+\frac{3}{2}\right|\le\frac{1}{4}\)
Dấu = xảy ra khi: \(x+\frac{3}{2}=0\Rightarrow x=-\frac{3}{2}\)
b) \(\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\)
Có: \(\left|x-\frac{4}{3}\right|\ge0;\left|y+\frac{1}{2}\right|\ge0\Rightarrow\frac{5}{3}-\left|x-\frac{4}{3}\right|-\left|y+\frac{1}{2}\right|\le\frac{5}{3}\)
Dấu = xảy ra khi: \(\) \(x-\frac{4}{3}=0\Rightarrow x=\frac{4}{3}\)
\(y+\frac{1}{2}=0\Rightarrow x=-\frac{1}{2}\)
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