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1. 计算:
(1)$-5^{2}+2×(-3)^{2}-(-6)÷(-\dfrac{1}{3})^{2}$;(2)$-1×[-3^{2}×(-\dfrac{2}{3})^{2}-2]÷(-\dfrac{2}{3})$.
(1)$-5^{2}+2×(-3)^{2}-(-6)÷(-\dfrac{1}{3})^{2}$;(2)$-1×[-3^{2}×(-\dfrac{2}{3})^{2}-2]÷(-\dfrac{2}{3})$.
答案
(1)47 (2)-9
2. 解下列方程：
(1) $1 - \dfrac{3 - 5x}{3} = \dfrac{3x - 5}{2}$；
(2) $\dfrac{|x| - 1}{5} - 1 = \dfrac{6 - |x|}{5}$。
(1) $1 - \dfrac{3 - 5x}{3} = \dfrac{3x - 5}{2}$；
(2) $\dfrac{|x| - 1}{5} - 1 = \dfrac{6 - |x|}{5}$。
答案
(1)$x=-15$ (2)$x=6$或$-6$
3. 先化简,再求值:
$5(3a^2b-ab^2-1)-(ab^2+3a^2b-5)$,其中$a=-\dfrac{1}{2},b=\dfrac{1}{3}$.
$5(3a^2b-ab^2-1)-(ab^2+3a^2b-5)$,其中$a=-\dfrac{1}{2},b=\dfrac{1}{3}$.
答案
$5(3a^{2}b-ab^{2}-1)-(ab^{2}+3a^{2}b-5)=15a^{2}b-5ab^{2}-5-ab^{2}-3a^{2}b+5=12a^{2}b-6ab^{2}.$
当$a=-\dfrac{1}{2},b=\dfrac{1}{3}$时,原式$=12× (-\dfrac{1}{2})^{2}× \dfrac{1}{3}-6× (-\dfrac{1}{2})× (\dfrac{1}{3})^{2}=1+\dfrac{1}{3}=\dfrac{4}{3}.$
当$a=-\dfrac{1}{2},b=\dfrac{1}{3}$时,原式$=12× (-\dfrac{1}{2})^{2}× \dfrac{1}{3}-6× (-\dfrac{1}{2})× (\dfrac{1}{3})^{2}=1+\dfrac{1}{3}=\dfrac{4}{3}.$
4. 先化简,再求值:
已知多项式$A=3a^2 - 6ab + b^2$,$B=-2a^2 + 3ab - 5b^2$,当$a=1,b=-1$时,试求$A+2B$的值.
已知多项式$A=3a^2 - 6ab + b^2$,$B=-2a^2 + 3ab - 5b^2$,当$a=1,b=-1$时,试求$A+2B$的值.
答案
因为$A=3a^{2}-6ab+b^{2},B=-2a^{2}+3ab-5b^{2}$,所以$A+2B=3a^{2}-6ab+b^{2}+2(-2a^{2}+3ab-5b^{2})=3a^{2}-6ab+b^{2}-4a^{2}+6ab-10b^{2}=-a^{2}-9b^{2}$,
当$a=1,b=-1$时,原式$=-1^{2}-9× (-1)^{2}=-10.$
当$a=1,b=-1$时,原式$=-1^{2}-9× (-1)^{2}=-10.$
5. 已知关于$x,y$的多项式$6mx^2 + 4nxy + 2x + 2xy - x^2 + y + 4$不含二次项，求$6m - 2n + 2$的值。
答案
原式$=(6m-1)x^{2}+(4n+2)xy+2x+y+4$,由多项式不含二次项,得到$6m-1=0,4n+2=0$,
解得$m=\dfrac{1}{6},n=-\dfrac{1}{2}$,则$6m-2n+2=1+1+2=4.$
解得$m=\dfrac{1}{6},n=-\dfrac{1}{2}$,则$6m-2n+2=1+1+2=4.$
