Question

BT7: Cho hai đa thức\(G=\left[\dfrac{1}{2}ax-2\left(ax+3\right)\right]-\left(ax+3\right)\)và\(H=ax-2-\left[-\left(ax-1\right)+3\right]-4\)

Tính G+H và G-H

Answer 1

\(G=\left[\dfrac{1}{2}ax-2\left(ax+3\right)\right]-\left(ax+3\right)\)

\(=\dfrac{1}{2}ax-2ax-6-ax-3\)

\(=\left(\dfrac{1}{2}ax-2ax-ax\right)+\left(-6-3\right)\)

\(=-\dfrac{5}{2}ax-9\)

\(H=ax-2-\left[-\left(ax-1\right)+3\right]-4\)

\(=ax-2-\left(-ax+4\right)-4\)

\(=ax-2+ax-4-4\)

\(=2ax-10\)

\(G+H=-\dfrac{5}{2}ax-9+2ax-10=-\dfrac{1}{2}ax-19\)

\(G-H=\left(-\dfrac{5}{2}ax-9\right)-\left(2ax-10\right)\)

\(=-\dfrac{5}{2}ax-9-2ax+10\)

\(=-\dfrac{9}{2}ax+1\)