Question

Tính giá trị của A=\(x^3y+2x^2-3xy^2-6\)tại \(x=\frac{1}{2},y=-2\)

Answer 1

Thay \(x=\frac{1}{2};y=-2\)

\(\Rightarrow A=\left(\frac{1}{2}\right)^3.\left(-2\right)+2\left(\frac{1}{2}\right)^2-3.\frac{1}{2}\left(-2\right)^2-6\)

\(\Rightarrow A=\frac{1}{8}.\left(-2\right)+2.\frac{1}{4}-3.\frac{1}{2}.\left(-4\right)-6\)

\(\Rightarrow A=\frac{-1}{4}+\frac{1}{2}-3+6-6\)

\(\Rightarrow A=\frac{1}{4}-3\)

\(\Rightarrow A=\frac{-11}{4}\)

Vậy \(A=\frac{-11}{4}\)