Question

[Nguyễn Khánh Sơn]

Tính GT biểu thức:

\(B=3x^2-2y^3-6x^2y^2+xy\)

vs \(x=\frac{2}{3};y=\frac{1}{2}\)

Answer 1

[o0o I am a studious person o0o]

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

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

\(=\frac{3}{4}\)