x²y–[2xy–2（xy–2分之3x²y）+x²y²],其中x=3，y=–1/3解：x²y–[2xy–2（xy–3x²y/2）+x²y²], = x²y–[2xy–2xy+3x²y+x²y²],= x²y–（3x²y+x²y²）= x²y–3x²y-x²y²=- x²y（2+y ） [其中x=3，y=–1/3]=-9×（–1/3）×（2–1/3）=5
