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a) có nghĩa khi \(x-1\ne0\Rightarrow x\ne1\)
b)\(f\left(7\right)=\frac{7+2}{7-1}=\frac{9}{6}\)
c)\(f\left(x\right)=\frac{x+2}{x-1}=\frac{1}{4}\Leftrightarrow x+2=4x-4\)
\(\Leftrightarrow-3x=-6\Leftrightarrow x=2\)
e)\(f\left(x\right)>1\Rightarrow\frac{x+2}{x-1}-1>0\)
\(\Rightarrow\frac{3}{x-1}>0\) thấy 3>0 nên x-1>0 =>x>1
Bài 2:
a)\(P=9-2\left|x-3\right|\)
Thấy: \(\left|x-3\right|\ge0\)\(\Rightarrow2\left|x-3\right|\ge0\)
\(\Rightarrow-2\left|x-3\right|\le0\)
\(\Rightarrow9-2\left|x-3\right|\le9\)
Khi x=3
b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(Q=\left|x-2\right|+\left|x-8\right|\)
\(=\left|x-2\right|+\left|8-x\right|\)
\(\ge\left|x-2+8-x\right|=6\)
Khi \(2\le x\le8\)
\(\left|x-4\right|+\left|x-5\right|+\left|x-6\right|\)
\(=\left|x-4\right|+\left|6-x\right|+\left|x-5\right|\)
\(\ge\left|x-4+6-x\right|+\left|x-5\right|=2+\left|x-5\right|\ge2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}\left(x-4\right)\left(6-x\right)\ge0\\x-5=0\end{matrix}\right.\)
\(\Rightarrow x=5\)
\(\left|x-5\right|+\left|x-7\right|\\ =\left|5-x\right|+\left|x-7\right|\\ \ge\left|5-x+x-7\right|\\ =\left|-2\right|\\ =2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(5-x\right)\left(x-7\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}5-x\ge0\\x-7\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}5-x\le0\\x-7\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le5\\x\ge7\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge5\\x\le7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow5\le x\le7\)
Vậy \(5\le x\le7\) thì \(\left|x-5\right|+\left|x-7\right|\) đạt GTNN