Sắp hết hạn rồi đấy ! ngày 10-1 mình sẽ giải .nhanh lên

\(p=a-\left\{\left(a-3\right)-\left[\left(a+3\right)-\left(-a-2\right)\right]\right\}\)

\(=a-\left\{a-3-\left[a+3+a+2\right]\right\}\)

\(=a-\left\{a-3-a-3-a-2\right\}\)

\(=a-\left\{-a-8\right\}\)

\(=a+a+8\)

\(=2a+8\)

\(Q=\left[a+\left(a+3\right)\right]-\left[\left(a+2\right)-\left(a-2\right)\right]\)

\(=\left[a+a+3\right]-\left[a+2-a+2\right]\)

\(=2a+3-4\)

\(=2a-1\)

Xét hiệu \(P-Q=\left(2a+8\right)-\left(2a-1\right)\)

\(=2a+8-2a+1\)

\(=9>0\)

Vậy: \(P>Q\)

Sáng mai mình gửi nha bạn

https://hoc24.vn/hoi-dap/question/165385.html

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 \(P>Q\)

nhanh lên nha

\(P>Q\)

P = a-[a-3-(a+3+a+2)]

   = a-[a-3-(2a+5)]

   = a-(a-3-2a-5)

   = a-(-a-8)

   = a+a+8 = 2a+8

Q = 2a+3-(a+2-a+2)

    = 2a+3-4 = 2a-1

Vì 2a+8 > 2a-1 => P > Q 

k mk nha

thank you bạn nha

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