Question

cho đa thức \(Q\left(x\right)=x\left(\dfrac{x^2}{2}-\dfrac{1}{2}x^3+\dfrac{1}{2}x\right)-\left(-\dfrac{1}{2}x^4+x^2\right)\)

tìm bậc của đa thức Q(x)

tính Q(-1/2)

Answer 1

Nhiều nick nhỉ! :)

Answer 2

ai giúp cho 10 like

Answer 3

a)\(x\left(\dfrac{x^2}{2}-\dfrac{1}{2}x^3+\dfrac{1}{2}x\right)-\left(-\dfrac{1}{2}x^4+x^2\right)\)

\(\Leftrightarrow\dfrac{x^3}{2}-\dfrac{1}{2}x^4+\dfrac{1}{2}x^2-\dfrac{1}{2}x^4-x^2\)

\(\Leftrightarrow\dfrac{1}{2}x^2\left(x-x^2+1\right)-x^2\)

vậy bậc của đa thức là 2

b)\(Q\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\left(-\dfrac{1}{2}\right)^2\left[-\dfrac{1}{2}-\left(-\dfrac{1}{2}\right)^2+1\right]-\left(-\dfrac{1}{2}\right)^2\)

\(=\dfrac{1}{2}.\dfrac{1}{4}\left(-\dfrac{1}{2}-\dfrac{1}{4}+1\right)-\dfrac{1}{4}=\dfrac{1}{8}.\dfrac{1}{4}-\dfrac{1}{4}\)

\(=\dfrac{1}{32}-\dfrac{1}{4}=-\dfrac{7}{32}\)