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 解：原式$=(5xy-\frac{9}{4}xy-\frac{1}{4}xy)+(-\frac{9}{2}x^{3}y^{2}+\frac{1}{2}x^{3}y^{2})-x^{3}y$

 $=(5 - \frac{9}{4}-\frac{1}{4})xy+(-\frac{9}{2}+\frac{1}{2})x^{3}y^{2}-x^{3}y$

 $=\frac{5}{2}xy - 4x^{3}y^{2}-x^{3}y$

 解：原式$=2(m - n)^{2}-(m - n)^{2}-(m - n)^{2}+(-1 + 4)(m - n)^{3}$

 $=(2 - 1 - 1)(m - n)^{2}+3(m - n)^{3}$

 $=3(m - n)^{3}$

 解：原式$=3x^{2}y^{2}-7x^{2}y^{2}+4x^{2}y^{2}+2xy-\frac{3}{2}xy + 2$

 $=(3 - 7 + 4)x^{2}y^{2}+(2-\frac{3}{2})xy+2$

 $=0\times x^{2}y^{2}+\frac{1}{2}xy + 2$

 $=\frac{1}{2}xy+2$

 把$x = 2，$$y=\frac{1}{4}$代入原式，得：

 $\frac{1}{2}\times2\times\frac{1}{4}+2$

 $=\frac{1}{4}+2$

 $=\frac{9}{4}$

 解：因?yàn)?3x^{a+1}y^{b-1}$與$\frac{2}{5}x^{2}y$是同類項(xiàng)，所以相同字母的指數(shù)相同，即$a + 1=2，$$b - 1=1。$解得$a=1，$$b=2。$

 原式$2a^{2}b + 3a^{2}b-\frac{1}{2}a^{2}b，$合并同類項(xiàng)可得：

 $\begin{aligned}&(2 + 3-\frac{1}{2})a^{2}b\\=&(5-\frac{1}{2})a^{2}b\\=&\frac{9}{2}a^{2}b\end{aligned}$

 把$a = 1，$$b = 2$代入$\frac{9}{2}a^{2}b，$得：

 $\frac{9}{2}\times1^{2}\times2=\frac{9}{2}\times1\times2 = 9$

 所以，原式的值為$9。$

 因?yàn)槎囗?xiàng)式不含$x^3$和$x^2$的項(xiàng)，所以$x^3$和$x^2$項(xiàng)的系數(shù)為$0。$

 對(duì)于$x^3$項(xiàng)：$m - 2 = 0，$解得$m = 2。$

 對(duì)于$x^2$項(xiàng)：$2(n + 1) = 0，$解得$n = -1。$

 將$m = 2，$$n = -1$代入多項(xiàng)式，得：

 $2x^4 + (2 - 2)x^3 + 2(-1 + 1)x^2 + 3x + \frac{-1}{2} = 2x^4 + 3x - \frac{1}{2}。$

 當(dāng)$x = -1$時(shí)，多項(xiàng)式的值為：

 $2×(-1)^4 + 3×(-1) - \frac{1}{2} = 2×1 - 3 - \frac{1}{2} = 2 - 3 - \frac{1}{2} = -1 - \frac{1}{2} = -\frac{3}{2}。$

 綜上，這個(gè)多項(xiàng)式為$2x^4 + 3x - \frac{1}{2}，$當(dāng)$x = -1$時(shí)該多項(xiàng)式的值為$-\frac{3}{2}。$

 解：對(duì)代數(shù)式$-5 + \frac{2}{3}x^2 - 3x + 4 + \frac{4}{3}x^2 + \frac{1}{2}x + 2 - 2x^2 + \frac{5}{2}x$進(jìn)行化簡(jiǎn)：

 $\begin{aligned}&(-5 + 4 + 2) + (\frac{2}{3}x^2 + \frac{4}{3}x^2 - 2x^2) + (-3x + \frac{1}{2}x + \frac{5}{2}x)\\=&1 + (\frac{6}{3}x^2 - 2x^2) + (-3x + 3x)\\=&1 + (2x^2 - 2x^2) + 0\\=&1\end{aligned}$

 化簡(jiǎn)后的結(jié)果為$1，$不含有$x，$所以無論$x$取何值，代數(shù)式的值都為$1，$因此小明和小亮選擇不同的$x$值代入計(jì)算，結(jié)果相同。

 解：因?yàn)?(a + 1)^2 + |b - 2| = 0，$且$(a + 1)^2 \geq 0，$$|b - 2| \geq 0，$所以$a + 1 = 0，$$b - 2 = 0，$解得$a = -1，$$b = 2。$

 對(duì)代數(shù)式$a^2b^2 + 3ab - 7a^2b^2 - \frac{5}{2}ab + 1 + 5a^2b^2$進(jìn)行化簡(jiǎn)：

 $\begin{aligned}&a^2b^2 + 3ab - 7a^2b^2 - \frac{5}{2}ab + 1 + 5a^2b^2\\=&(a^2b^2 - 7a^2b^2 + 5a^2b^2) + (3ab - \frac{5}{2}ab) + 1\\=&(-a^2b^2) + (\frac{6}{2}ab - \frac{5}{2}ab) + 1\\=&-a^2b^2 + \frac{1}{2}ab + 1\end{aligned}$

 將$a = -1，$$b = 2$代入化簡(jiǎn)后的式子：

 $\begin{aligned}&-(-1)^2 \times 2^2 + \frac{1}{2} \times (-1) \times 2 + 1\\=&-1 \times 4 + (-1) + 1\\=&-4 - 1 + 1\\=&-4\end{aligned}$

 故代數(shù)式的值為$-4。$

【答案】：
???
解:?原式$=\frac {1}{2}xy+2??$
??把$x=2,y=\frac {1}{4}$代入原式$=\frac {1}{2}×2×\frac {1}{4}+2=\frac {9}{4}??$

【解析】：
$3x^{2}y^{2}+2xy - 7x^{2}y^{2}-\frac{3}{2}xy + 2 + 4x^{2}y^{2}$
$=(3x^{2}y^{2}-7x^{2}y^{2}+4x^{2}y^{2})+(2xy-\frac{3}{2}xy)+2$
$=\frac{1}{2}xy+2$
當(dāng)$x = 2$，$y= \frac{1}{4}$時(shí)，
$\frac{1}{2}xy+2=\frac{1}{2}×2×\frac{1}{4}+2=\frac{1}{4}+2=\frac{9}{4}$
$\frac{9}{4}$

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