初中数学七年级上册（581题）

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若关于\(x\)、\(y\)的多项式\(6m{x^2} + 4nxy + 2x + 2xy - {x^2} + y + 4\)不含二次项，\(m - n\)的值。

知识点：第二章　整式的加减

参考答案：原式\( = (6m - 1){x^2} + (4n + 2)xy + 2x + y + 4\)
\(\because \)多项式不含二次项
\(\therefore 6m - 1 = 0\)，
\(4n + 2 = 0\)
\(\therefore \)\(m = \frac{1}{6},n = - \frac{1}{2}\)
\(\therefore \)\(m - n = \frac{1}{6} - ( - \frac{1}{2}) = \frac{2}{3}\)