\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(=\left(a-3b+a+3b\right)\left(a-3b-a-3b\right)-\left(ab-2a-b+2\right)\)
\(=2a.\left(-6b\right)-ab+2a+b-2\)
\(=-12ab-ab+2a+b-2\)
\(=-13ab+2a+b-2\)
Thay a=1/2,b=-3 vào N ta được:
\(N=-13\cdot\frac{1}{2}.\left(-3\right)+2\cdot\frac{1}{2}-3-2=\frac{39}{2}+\frac{2}{2}-5=\frac{41}{2}-5=\frac{31}{2}\)
\(N=\left(a-3b\right)^2-\left(a+3b\right)^2-\left(a-1\right)\left(b-2\right)\)
\(\Rightarrow N=\left(a-3b\right)^2-\left[\left(a-3b\right)+6b\right]^2-\left(ab-2a-b+2\right)\)
\(\Rightarrow N=\left(a-3b\right)^2-\left(a-3b\right)^2-2\left(a-3b\right)6b-36b^2-ab+2a-b+2\)
\(\Rightarrow N=\left(-2a+6b\right)6b-36b^2-ab+2a-b+2\)
\(\Rightarrow N=-12a+36b^2-36b^2-ab+2a-b+2\)
\(\Rightarrow N=-10a-ab-b+2\)
Thay a = 1/2 ; b = -3 vào ta có
\(\Rightarrow N=-10.\frac{1}{2}-\frac{1}{2}.-3+3+2\)
\(\Leftrightarrow N=-5+\frac{3}{2}+5=\frac{3}{2}\)
