\(A=2x-3;B=6-x.\)
a) TH1: \(A=B.\)
\(\Rightarrow2x-3=6-x\)
\(\Rightarrow2x+x=6+3\)
\(\Rightarrow3x=9\)
\(\Rightarrow x=9:3\)
\(\Rightarrow x=3.\)
+ TH2: \(A>B.\)
\(\Rightarrow2x-3>6-x\)
\(\Rightarrow2x+x>6+3\)
\(\Rightarrow3x>9\)
\(\Rightarrow x>9:3\)
\(\Rightarrow x>3.\)
+ TH3: \(A< B.\)
\(\Rightarrow2x-3< 6-x\)
\(\Rightarrow2x+x< 6+3\)
\(\Rightarrow3x< 9\)
\(\Rightarrow x< 9:3\)
\(\Rightarrow x< 3.\)
Vậy khi \(x=3\) thì \(A=B.\)
khi \(x>3\) thì \(A>B.\)
khi \(x< 3\) thì \(A< B.\)
b) Để tích \(A.B\) có giá trị dương.
\(\Leftrightarrow A.B>0\)
\(\Rightarrow\left(2x-3\right).\left(6-x\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3>0\\6-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3< 0\\6-x< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x>3\\x< 6\end{matrix}\right.\\\left\{{}\begin{matrix}2x< 3\\x>6\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\frac{3}{2}\\x< 6\end{matrix}\right.\\\left\{{}\begin{matrix}x< \frac{3}{2}\\x>6\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{2}< x< 6\\x\in\varnothing\end{matrix}\right.\)
Vậy nếu \(\frac{3}{2}< x< 6\) thì tích \(A.B\) có giá trị dương.
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