Question

a) [ x . (x+4) . (x-4) - (x\(^2\)+1)]. x\(^2\) - 1

b) (y-3). y+3 . y\(^2\)+9 -y\(^2\)+2 . ( y\(^2\)-2)

c) (x+y). (x\(^2\)x\(^{^{ }2}\)- xy+ y\(^{^{ }2}\))

d) (\(\dfrac{1}{2}\) xy + \(\dfrac{3}{4}\)y ) . \(\dfrac{1}{2}\) xy - \(\dfrac{3}{4}\) y

Answer 1

a, \(\left[x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\right]x^2-1\)

\(=\left[x\left(x^2-16\right)-\left(x^2+1\right)\right]x^2-1\)

\(=\left[x^3-16x-x^2-1\right]x^2-1\)

\(=x^5-16x^3-x^4-x^2-1\)

b, \(\left(y-3\right)y+3y^2+9-y^2+2\left(y^2-2\right)\)

\(=y^2-3y+3y^2+9-y^2+2y^2-4\)

\(=5y^2-3y+5\)

c, \(\left(x+y\right)\left(x^2x^2-xy+y^2\right)\)

\(=x^5-x^2y+xy^2+x^4y-xy^2+y^3\)

d, \(\left(\dfrac{1}{2}xy+\dfrac{3}{4}y\right).\dfrac{1}{2}xy-\dfrac{3}{4}y\)

\(=\dfrac{1}{4}x^2y^2+\dfrac{3}{8}xy^2-\dfrac{3}{4}y\)

\(=\dfrac{1}{4}y.\left(x^2y+\dfrac{3}{2}xy-3\right)\)

Chúc bạn học tốt!!!