Bài 1 :
a ) \(A=\left|2,5-x\right|+5,8\)
Ta có : \(\left|2,5-x\right|\ge0\)
\(\Leftrightarrow\left|2,5-x\right|+5,8\ge5,8\)
Vậy \(GTNN\) là 5,8 khi và chỉ khi \(x=2,5.\)
b ) \(B=2-\left|x+\dfrac{2}{3}\right|\)
Ta có : \(\left|x+\dfrac{2}{3}\right|\ge0\)
\(\Leftrightarrow2-\left|x+\dfrac{2}{3}\right|\le2\)
Vậy \(GTLN\) là 2 khi và chỉ khi \(x=-\dfrac{2}{3}.\)
1,
a, Ta có: \(\left|2,5-x\right|\ge0\Rightarrow A=\left|2,5-x\right|+5,8\ge5,8\)
Dấu " = " khi \(\left|2,5-x\right|=0\Rightarrow x=2,5\)
Vậy \(MIN_A=5,8\) khi x = 2,5
b, Ta có: \(-\left|x+\dfrac{2}{3}\right|\le0\Rightarrow B=2-\left|x+\dfrac{2}{3}\right|\le2\)
Dấu " = " khi \(-\left|x+\dfrac{3}{2}\right|=0\Rightarrow x=\dfrac{-3}{2}\)
Vậy \(MAX_B=2\) khi \(x=\dfrac{-3}{2}\)
2,
Giải:
Oy // AC \(\Rightarrow\widehat{O_1}+\widehat{C_2}=180^o\) ( 2 góc trong cùng phía )
\(\Rightarrow\widehat{C_2}=110^o\left(\widehat{O_1}=70^o\right)\)
AB // Ox \(\Rightarrow\widehat{A}+\widehat{C_2}=180^o\) ( 2 góc trong cùng phía )
\(\Rightarrow\widehat{A}=70^o\left(\widehat{C_2}=110^o\right)\)
Vậy...
\(A=\left|2,5-x\right|+5,8\)
\(\left|2,5-x\right|\ge0\)
\(A_{MIN}\Rightarrow\left|2,5-x\right|_{MIN}\)
\(\left|2,5-x\right|_{MIN}=0\Rightarrow x=2,5\)
\(A_{MIN}=0+5,8=5,8\)
\(\Rightarrow A_{MIN}=5,8\) khi \(x=2,5\)
\(B=2-\left|x+\dfrac{2}{3}\right|\)
\(\left|x+\dfrac{2}{3}\right|\ge0\)
\(B_{MAX}\Rightarrow\left|x+\dfrac{2}{3}\right|_{MIN}\)
\(\left|x+\dfrac{2}{3}\right|_{MIN}=0\Rightarrow x=-\dfrac{2}{3}\)
\(\Rightarrow B_{MAX}=2-0=0\) khi \(x=-\dfrac{2}{3}\)