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1. 计算:
(1) $(-\dfrac{4}{5})×13 + (-\dfrac{4}{5})×2 - (-\dfrac{4}{5})×5$；
(2) $(-2)^4 ÷ (-2\dfrac{2}{3})^2 + 5\dfrac{1}{2} × (-\dfrac{1}{6}) - 0.25$。
(1) $(-\dfrac{4}{5})×13 + (-\dfrac{4}{5})×2 - (-\dfrac{4}{5})×5$；
(2) $(-2)^4 ÷ (-2\dfrac{2}{3})^2 + 5\dfrac{1}{2} × (-\dfrac{1}{6}) - 0.25$。
答案
(1)-8
(2)$1\dfrac{1}{12}$
(2)$1\dfrac{1}{12}$
2. 解下列方程：
(1)$\dfrac{7x - 1}{3} - \dfrac{5x + 1}{2} = 1 - \dfrac{3x + 2}{4};$
(2)$\dfrac{0.03x - 0.1}{0.02} = \dfrac{0.1x + 0.2}{0.3} - 2.$
(1)$\dfrac{7x - 1}{3} - \dfrac{5x + 1}{2} = 1 - \dfrac{3x + 2}{4};$
(2)$\dfrac{0.03x - 0.1}{0.02} = \dfrac{0.1x + 0.2}{0.3} - 2.$
答案
(1)$x=\dfrac{16}{7}$
(2)$x=\dfrac{22}{7}$
(2)$x=\dfrac{22}{7}$
3. 先化简,再求值:
$\frac{1}{2}x - 2(x - \frac{1}{3}y^2) + (-\frac{3}{2}x + \frac{1}{3}y^2)$,其中$x,y$满足$|x - 2| + (y + 1)^2 = 0$.
$\frac{1}{2}x - 2(x - \frac{1}{3}y^2) + (-\frac{3}{2}x + \frac{1}{3}y^2)$,其中$x,y$满足$|x - 2| + (y + 1)^2 = 0$.
答案
由题意得,$x-2=0,y+1=0$,解得$x=2,y=-1$,原式=$\frac{1}{2}x-2x+\frac{2}{3}y^2-\frac{3}{2}x+\frac{1}{3}y^2=-3x+y^2$,当$x=2,y=-1$时,原式=$-3×2+1=-5.$
4. 先化简,再求值:
$\frac{1}{2}a^2b-[\frac{3}{2}a^2b-(3abc-a^2c)-4a^2c]-3abc$,其中关于$x,y$的单项式$cx^{2a+2}y^2$与$0.4xy^{3b+4}$的和为零.
$\frac{1}{2}a^2b-[\frac{3}{2}a^2b-(3abc-a^2c)-4a^2c]-3abc$,其中关于$x,y$的单项式$cx^{2a+2}y^2$与$0.4xy^{3b+4}$的和为零.
答案
$a=-\dfrac{1}{2},b=-\dfrac{2}{3},c=-0.4$,原式=$-a^2b+3a^2c=-\dfrac{2}{15}.$
5. 已知$A=x^3-2y^3+3x^2y+xy^2-3xy+4$，$B=y^3-x^3-4x^2y-3xy-3xy^2+3$，$C=y^3+x^2y+2xy^2+6xy-6$。试说明无论$x,y$取何值，$A+B+C$都是常数。
答案
$A+B+C=x^3-2y^3+3x^2y+xy^2-3xy+4+y^3-x^3-4x^2y-3xy-3xy^2+3+y^3+x^2y+2xy^2+6xy-6=x^3-x^3-2y^3+y^3+y^3+3x^2y-4x^2y+x^2y+xy^2-3xy^2+2xy^2+6xy-3xy-3xy-6+4+3=1$,所以无论$x,y$取何值,$A+B+C$都是常数.
