+)Ta có :\(P=a-\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(-a-2\right)\right]\right\}\)
\(\Rightarrow P=a-\left\{a-3-\left(a+3-\left(-a\right)+2\right)\right\}\)
\(\Rightarrow P=a-\left\{a-3-\left(a+3+a+2\right)\right\}\)
\(\Rightarrow P=a-\left\{a-3-a-3-a-2\right\}\)
\(\Rightarrow P=a+a+3+a+3+a+2\)
\(\Rightarrow P=4a+8\)
+)Ta lại có:\(Q=\left[a+\left(a+3\right)\right]-\left[\left(a+2\right)-\left(a-2\right)\right]\)
\(\Rightarrow Q=\left(a+a+3\right)-\left(a+2-a+2\right)\)
\(\Rightarrow Q=2a+3-4\)
\(\Rightarrow Q=2a-1\)
+)Ta thấy 4a+8>2a-1
\(\Rightarrow P>Q\)
Vậy \(P>Q\)
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