Question

cho 2 đa thức M =-xy^2+3x^2y -x^2y^2

N=1/2x2y-xy^2 + -2/3x^2y^2

a.Tính M+ N

b.Tìm Q biết N-Q=M

c ,Tính giá trị đa thức Q tại x=-1 y=1/2

Answer 1

a: Ta có: M+N

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

\(=-2xy^2+\dfrac{7}{2}x^2y-\dfrac{5}{3}x^2y^2\)

b: Ta có: N-Q=M

nên \(Q=N-M\)

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

\(=\dfrac{-5}{2}x^2y+\dfrac{1}{3}x^2y^2\)

Answer 2

a) \(M+N=-xy^2+3x^2y-x^2y^2+\dfrac{1}{2}x^2y-xy^2-\dfrac{2}{3}x^2y^2=\dfrac{7}{2}x^2y-2xy^2-\dfrac{5}{3}x^2y^2\)b) \(N-Q=M\Rightarrow Q=N-M=\dfrac{1}{2}x^2y-xy^2-\dfrac{2}{3}x^2y^2+xy^2-3x^2y+x^2y^2=-\dfrac{5}{2}x^2y+\dfrac{1}{3}x^2y^2\)c) \(Q=-\dfrac{5}{2}x^2y+\dfrac{1}{3}x^2y^2=-\dfrac{5}{2}.\left(-1\right)^2.\dfrac{1}{2}+\dfrac{1}{3}.\left(-1\right)^2.\left(\dfrac{1}{2}\right)^2=-\dfrac{7}{6}\)

Answer 3

c: Thay x=-1 và \(y=\dfrac{1}{2}\) vào Q, ta được:

\(Q=-\dfrac{5}{2}\cdot1\cdot\dfrac{1}{2}+\dfrac{1}{3}\cdot1\cdot\dfrac{1}{4}\)

\(=-\dfrac{5}{4}+\dfrac{1}{12}\)

\(=-\dfrac{15}{12}+\dfrac{1}{12}=-\dfrac{14}{12}=-\dfrac{7}{6}\)