Question

[Phuong Anh Do]

Rút gọn và tính giá trị biểu thức :  

 \(N=\) \(\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)\(với\)\(a=\dfrac{1}{2};b=-3\)

Answer 1

[Lấp La Lấp Lánh]

\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)=\left(a-3b-a-3b\right)\left(a-3b+a+3b\right)-\left(ab-2a-b+2\right)=\left(-6b\right).2a-ab+2a+b-2=2a+b-13ab-2\)

Thay \(a=\dfrac{1}{2};b=-3\) vào N ta được: \(N=2a+b-13ab-2=2.\dfrac{1}{2}-3-13.\dfrac{1}{2}.\left(-3\right)-2=\dfrac{31}{2}\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Ta có: \(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)

\(=a^2-6ab+9b^2-a^2-6ab-9b^2-ab+2a+b-2\)

\(=-13ab+2a+b-2\)

\(=-13\cdot\dfrac{1}{2}\cdot\left(-3\right)-1-3-2\)

\(=\dfrac{27}{2}\)