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![](https://rs.olm.vn/images/avt/0.png?1311)
\(p=a-\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(-a-2\right)\right]\right\}\)
\(=a-\left\{a-3-\left[a+3+a+2\right]\right\}\)
\(=a-\left\{a-3-a-3-a-2\right\}\)
\(=a-\left\{-a-8\right\}\)
\(=a+a+8\)
\(=2a+8\)
\(Q=\left[a+\left(a+3\right)\right]-\left[\left(a+2\right)-\left(a-2\right)\right]\)
\(=\left[a+a+3\right]-\left[a+2-a+2\right]\)
\(=2a+3-4\)
\(=2a-1\)
Xét hiệu \(P-Q=\left(2a+8\right)-\left(2a-1\right)\)
\(=2a+8-2a+1\)
\(=9>0\)
Vậy: \(P>Q\)
![](https://rs.olm.vn/images/avt/0.png?1311)
P = a – {(a – 3) – [(a + 3) – (- a – 2)]
= a – {a – 3 – [a + 3 + a + 2]} = a – {a – 3 – a – 3 – a – 2}
= a – {- a – 8} = a + a + 8 = 2a + 8.
Q = [a+ (a + 3)] – [a + 2 – (a – 2)]
= [a + a + 3] – [a + 2 – a + 2] = 2a + 3 – 4 = 2a – 1
Xét hiệu P – Q = (2a + 8) – (2a – 1) = 2a + 8 – 2a + 1 = 9 > 0
Vậy P > Q
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
P=a-{(a-3)-[(a+3)-(-a-2)]}
P=a-{a-3-[a+3-(-a)+2]}
P=a-{a-3-[a+3+a+2]}
P=a-{a-3-[2a+5]}
P=a-{a-3-2a-5}
P=a-{(-a)-8}
P=a-(-a)+8
P=a+a+8
P=2a+8
Q=[a+(a+3)]-{(a+2)-(a-2)]
Q=[a+a+3]-[a+2-a+2]
Q=[2a+3]-4
Q=2a+3-4
Q=2a+(-1)
Ta có:P=2a+8
Q=2a+(-1)
=>P>Q
Chúc bn học tốt
P = a - {(a - 3) - [(a + 3) - (-a - 2)]}
= a - {a - 3 - [a + 3 - (-a) + 2]}
= a - {a - 3 - a - 3 + (-a) - 2}
= a - a + 3 + a + 3 - (-a) + 2
= a + (-a) + 3 + a + 3 + a + 2
= [a + (-a) + a + a] + (3 + 3 + 2)
= 2a +8
Q = [a + (a + 3)] - [(a + 2) - (a - 2)]
= [a + a + 3] - [a + 2 - a + 2]
= a + a + 3 - a - 2 + a - 2
= a + a + 3 + (-a) + (-2) + a + (-2)
= [a + a + (-a) + a] + [3 + (-2) + (-2)]
= 2a + (-1)
=> 8 > -1
=> 2a + 8 > 2a + (-1)
=> P > Q
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=a.\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(-a-2\right)\right]\right\}.\)
\(P=a.\left\{\left(a-3\right)-\left[a+3+a+2\right]\right\}\)
\(P=a\left[\left(a-3-2a+5\right)\right]\)
\(P=a\left(2-a\right)\)
\(P=2a-a^2_{\left(1\right)}\)
\(Q=\left[a+\left(a+3\right)\right]-\left[\left(a+2\right)-\left(a-2\right)\right]\)
\(Q=\left(a+a+3\right)-\left(a+2-a+2\right)\)
\(Q=2a+3-4\)
\(Q=2a-1_{\left(2\right)}\)
\(TH1:\) Với \(a=0\)
thì P>Q
\(TH2:\) Với \(a\ne0\)
thì P<Q
![](https://rs.olm.vn/images/avt/0.png?1311)
Theo bài ra ta có :
\(P=a-\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(a-2\right)\right]\right\}\)
\(=a-\left[a-3-\left(a+3-a+2\right)\right]\)
\(=a-\left(a-3-5\right)=a-\left(a-8\right)=8\)
\(Q=\left[a+\left(a+3\right)\right]-\left[a+2-\left(a-2\right)\right]\)
\(=\left(a+a+3\right)-\left(a+2-a+2\right)=\left(2a+3\right)-4=2a-1\)
Đặt \(2a-1=0\Leftrightarrow a=\frac{1}{2}\)Thay a = 1/2 ta được : \(2.\frac{1}{2}-1=1-1=0\)
mà \(8>0\)hay \(P>Q\)
P=a{(a-3)-(a+3)-(-a-2)
P=a[a-3-(a+3+a-2)
P=a(a-3-a-3-a-2)
P=a(-a-8)
P=a2-8a
Q=[a+(a+3)]-[(a+2)-(a-2)]
Q=(a+a+3)-(a+2-a+2)
Q=2a+3-4
Q=2a-1
=)P<Q