19.(6分)新素养 运算能力 计算：
(1)$5(2x - 7y)-3(4x - 10y)$；
(2)$\frac{1}{2}(a^{2}-b)+\frac{1}{3}(a - b^{2})+\frac{1}{6}(a^{2}+b^{2})$.
(1)$5(2x - 7y)-3(4x - 10y)$；
(2)$\frac{1}{2}(a^{2}-b)+\frac{1}{3}(a - b^{2})+\frac{1}{6}(a^{2}+b^{2})$.
答案：(1)原式$=10x-35y-12x+30y=-2x-5y.$(2)原式$=\frac {1}{2}a^{2}-\frac {1}{2}b+\frac {1}{3}a-\frac {1}{3}b^{2}+\frac {1}{6}a^{2}+\frac {1}{6}b^{2}=\frac {2}{3}a^{2}+\frac {1}{3}a-\frac {1}{6}b^{2}-\frac {1}{2}b.$
20.(4分)先化简,再求值：$2a^{2}-[\frac{1}{2}(ab - 4a^{2})-7ab]-\frac{1}{2}ab$,其中$a$,$b满足\vert a+\frac{1}{2}\vert+(b - 3)^{2}= 0$.
答案：原式$=2a^{2}-(\frac {1}{2}ab-2a^{2}-7ab)-\frac {1}{2}ab=2a^{2}-\frac {1}{2}ab+2a^{2}+7ab-\frac {1}{2}ab=4a^{2}+6ab$.由题意,得$a+\frac {1}{2}=0,b-3=0$,所以$a=-\frac {1}{2},b=3$,所以原式$=4×(-\frac {1}{2})^{2}+6×(-\frac {1}{2})×3=1-9=-8.$
解析：
解：原式$=2a^{2}-(\frac{1}{2}ab - 2a^{2}-7ab)-\frac{1}{2}ab$
$=2a^{2}-\frac{1}{2}ab + 2a^{2}+7ab-\frac{1}{2}ab$
$=4a^{2}+6ab$
由$\vert a+\frac{1}{2}\vert+(b - 3)^{2}= 0$，得$a+\frac{1}{2}=0$，$b - 3=0$，所以$a=-\frac{1}{2}$，$b=3$。
原式$=4×(-\frac{1}{2})^{2}+6×(-\frac{1}{2})×3$
$=4×\frac{1}{4}+(-9)$
$=1 - 9$
$=-8$
$=2a^{2}-\frac{1}{2}ab + 2a^{2}+7ab-\frac{1}{2}ab$
$=4a^{2}+6ab$
由$\vert a+\frac{1}{2}\vert+(b - 3)^{2}= 0$，得$a+\frac{1}{2}=0$，$b - 3=0$，所以$a=-\frac{1}{2}$，$b=3$。
原式$=4×(-\frac{1}{2})^{2}+6×(-\frac{1}{2})×3$
$=4×\frac{1}{4}+(-9)$
$=1 - 9$
$=-8$
21.(4分)(2025·江苏盐城期末)已知代数式$2x^{2}+ax - y + 6与2bx^{2}-3x + 5y - 1的差与x$的取值无关,求代数式$\frac{1}{3}a^{3}-3b^{2}-(\frac{1}{4}a^{3}-2b^{2})$的值.
答案：$(2x^{2}+ax-y+6)-(2bx^{2}-3x+5y-1)=2x^{2}+ax-y+6-2bx^{2}+3x-5y+1=(2-2b)x^{2}+(a+3)x-6y+7$.由题意,得$2-2b=0,a+3=0$,所以$b=1,a=-3$,所以$\frac {1}{3}a^{3}-3b^{2}-(\frac {1}{4}a^{3}-2b^{2})=\frac {1}{3}a^{3}-3b^{2}-\frac {1}{4}a^{3}+2b^{2}=\frac {1}{12}a^{3}-b^{2}=\frac {1}{12}×(-3)^{3}-1^{2}=-\frac {9}{4}-1=-\frac {13}{4}.$
解析：
$(2x^{2}+ax - y + 6)-(2bx^{2}-3x + 5y - 1)$
$=2x^{2}+ax - y + 6 - 2bx^{2}+3x - 5y + 1$
$=(2 - 2b)x^{2}+(a + 3)x - 6y + 7$
因为差与$x$的取值无关，所以$2 - 2b = 0$，$a + 3 = 0$，解得$b = 1$，$a = - 3$。
$\frac{1}{3}a^{3}-3b^{2}-(\frac{1}{4}a^{3}-2b^{2})$
$=\frac{1}{3}a^{3}-3b^{2}-\frac{1}{4}a^{3}+2b^{2}$
$=(\frac{1}{3}-\frac{1}{4})a^{3}+(-3b^{2}+2b^{2})$
$=\frac{1}{12}a^{3}-b^{2}$
将$a = - 3$，$b = 1$代入，得：
$\frac{1}{12}×(-3)^{3}-1^{2}=\frac{1}{12}×(-27)-1=-\frac{9}{4}-1=-\frac{13}{4}$
$-\frac{13}{4}$
$=2x^{2}+ax - y + 6 - 2bx^{2}+3x - 5y + 1$
$=(2 - 2b)x^{2}+(a + 3)x - 6y + 7$
因为差与$x$的取值无关，所以$2 - 2b = 0$，$a + 3 = 0$，解得$b = 1$，$a = - 3$。
$\frac{1}{3}a^{3}-3b^{2}-(\frac{1}{4}a^{3}-2b^{2})$
$=\frac{1}{3}a^{3}-3b^{2}-\frac{1}{4}a^{3}+2b^{2}$
$=(\frac{1}{3}-\frac{1}{4})a^{3}+(-3b^{2}+2b^{2})$
$=\frac{1}{12}a^{3}-b^{2}$
将$a = - 3$，$b = 1$代入，得：
$\frac{1}{12}×(-3)^{3}-1^{2}=\frac{1}{12}×(-27)-1=-\frac{9}{4}-1=-\frac{13}{4}$
$-\frac{13}{4}$