Question

Cho đa thức Q(x)=\(\dfrac{1}{6}\)xy²+\(\dfrac{1}{3}\)x-\(\dfrac{3}{4}\)xy²-1+x+\(\dfrac{1}{4}\)x²y

a, Tìm A để: A-Q=x²y-3x+1

b, Tìm P để: P-\(\dfrac{3}{4}\)xy²-\(\dfrac{4}{9}\)x=Q

c, Tìm M để: -xy²+\(\dfrac{4}{3}\)x-\(\dfrac{3}{4}\)-M=Q

Answer 1

a: \(Q=-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1\)

\(A=x^2y-3x+1-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{2}x^2y-\dfrac{7}{12}xy^2-3x\)

b: \(P=\dfrac{3}{4}xy^2+\dfrac{4}{9}x-\dfrac{7}{12}xy^2+\dfrac{4}{3}x-\dfrac{1}{2}x^2y-1=\dfrac{1}{6}xy^2+\dfrac{16}{9}x-\dfrac{1}{2}x^2y-1\)