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1. 先化简,再求值:$5(3a^{2}b-ab^{2})-2(-ab^{2}+3a^{2}b)$,其中$a=-2,b=-3.$
答案
原式=$15a^{2}b - 5ab^{2} + 2ab^{2} - 6a^{2}b = 9a^{2}b - 3ab^{2}$.当$a = -2,b = -3$时,原式=$9×(-2)^{2}×(-3) - 3×(-2)×(-3)^{2} = -108 + 54 = -54$.
2. 先化简,再求值: $(2x^{3}-2y^{2})-3(x^{3}y^{2}+x^{3})+2(y^{2}+y^{2}x^{3})$, 其中 $x=-1,y=2$.
答案
原式=$2x^{3} - 2y^{2} - 3x^{3}y^{2} - 3x^{3} + 2y^{2} + 2x^{3}y^{2} = -x^{3} - x^{3}y^{2}$.当$x = -1,y = 2$时,原式=$-(-1)^{3} - (-1)^{3}×2^{2} = 1 + 4 = 5$.
3. 已知$(x-3)^{2}+|y+2|=0$,求代数式$2x^{2}+(-x^{2}-2xy+2y^{2})-2(x^{2}-xy+2y^{2})$的值.
答案
原式=$2x^{2} - x^{2} - 2xy + 2y^{2} - 2x^{2} + 2xy - 4y^{2} = -x^{2} - 2y^{2}$.因为$(x - 3)^{2} + |y + 2| = 0$,所以$x - 3 = 0,y + 2 = 0$,所以$x = 3,y = -2$,所以原式=$-3^{2} - 2×(-2)^{2} = -9 - 8 = -17$.
4. 若多项式$(x^{2}-3kxy-3y^{2})+(\dfrac{1}{3}xy-8)$($k$ 为常数)中不含 $xy$ 项,求 $k$ 的值.
答案
原式=$x^{2} + (\dfrac{1}{3} - 3k)xy - 3y^{2} - 8$.因为该多项式中不含$xy$项,所以$\dfrac{1}{3} - 3k = 0$,解得$k = \dfrac{1}{9}$.
5. 已知 $A-B=7a^{2}-7ab$,其中 $B=-4a^{2}+6ab+7$,且 $|a+1|+(b-2)^{2}=0$.
(1)填空:$a=$
(2)求 $A$ 的值.
(1)填空:$a=$
-1
,$b=$2
.(2)求 $A$ 的值.
答案
(1)-1 2 解析:因为$|a + 1| + (b - 2)^{2} = 0$,所以$a + 1 = 0,b - 2 = 0$,解得$a = -1,b = 2$.
(2)因为$A - B = 7a^{2} - 7ab$,$B = -4a^{2} + 6ab + 7$,所以$A = (A - B) + B = 7a^{2} - 7ab + (-4a^{2} + 6ab + 7) = 3a^{2} - ab + 7$.当$a = -1,b = 2$时,$A = 3×(-1)^{2} - (-1)×2 + 7 = 3 + 2 + 7 = 12$.
(2)因为$A - B = 7a^{2} - 7ab$,$B = -4a^{2} + 6ab + 7$,所以$A = (A - B) + B = 7a^{2} - 7ab + (-4a^{2} + 6ab + 7) = 3a^{2} - ab + 7$.当$a = -1,b = 2$时,$A = 3×(-1)^{2} - (-1)×2 + 7 = 3 + 2 + 7 = 12$.
