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1. $2^{-3}$ 的值是(
A.-6
B.-8
C.$\frac{1}{8}$
D.$-\frac{1}{8}$
C
).A.-6
B.-8
C.$\frac{1}{8}$
D.$-\frac{1}{8}$
答案
C
2.计算：$(a^{-1}b^{2})^{3}=$
$\frac{b^6}{a^3}$
.答案
$\frac{b^6}{a^3}$
3.若$ a^{-1}=(-1)^0 $，则$ a= $
1
。答案
1
4. 计算:
(1)$(2m^{2}n^{-3})^{2}· (-mn^{-2})^{-2};$
(2)$4x^{2}y^{-3}z÷ (-2x^{-1}yz^{-2})^{2};$
(3)$(\frac{1}{2})^{-3}-2^{3}×0.125+2136^{0}+|-1|;$
(4)$2×(π-3.14)^{0}+8×(3×2)^{-1}+(-\frac{1}{2})^{-4}.$
(1)$(2m^{2}n^{-3})^{2}· (-mn^{-2})^{-2};$
(2)$4x^{2}y^{-3}z÷ (-2x^{-1}yz^{-2})^{2};$
(3)$(\frac{1}{2})^{-3}-2^{3}×0.125+2136^{0}+|-1|;$
(4)$2×(π-3.14)^{0}+8×(3×2)^{-1}+(-\frac{1}{2})^{-4}.$
答案
解 (1)原式$=4m^{4}n^{-6} · m^{-2}n^{4}$
$=4m^{2}n^{-2}=\dfrac{4m^{2}}{n^{2}}.$
(2)原式$=4x^{2}y^{-3}z÷ (4x^{-2}y^{2}z^{-4})$
$=x^{4}y^{-5}z^{5}=\dfrac{x^{4}z^{5}}{y^{5}}.$
(3)原式$=8-8×0.125+1+1$
$=8-1+1+1$
$=9.$
(4)原式$=2×1+8×\dfrac{1}{6}+16$
$=2+\dfrac{4}{3}+16$
$=19\dfrac{1}{3}.$
$=4m^{2}n^{-2}=\dfrac{4m^{2}}{n^{2}}.$
(2)原式$=4x^{2}y^{-3}z÷ (4x^{-2}y^{2}z^{-4})$
$=x^{4}y^{-5}z^{5}=\dfrac{x^{4}z^{5}}{y^{5}}.$
(3)原式$=8-8×0.125+1+1$
$=8-1+1+1$
$=9.$
(4)原式$=2×1+8×\dfrac{1}{6}+16$
$=2+\dfrac{4}{3}+16$
$=19\dfrac{1}{3}.$
5. 已知$|x+y-3|+(x-y-1)^2=0$，求$\frac{1}{2}[(-x^{-2}y)^2]^{-3}$的值.
答案
解 $\because |x+y-3|+(x-y-1)^2=0,$
$\therefore \begin{cases} x+y-3=0,\\ x-y-1=0. \end{cases}$
解得$\begin{cases} x=2,\\ y=1. \end{cases}$
$\therefore \dfrac{1}{2}[(-x^{-2}y)^{2}]^{-3}$
$=\dfrac{1}{2}(-x^{-2}y)^{-6}$
$=\dfrac{1}{2}(x^{12} · y^{-6})$
$=\dfrac{1}{2}×2^{12}×1^{-6}$
$=2^{11}.$
$\therefore \begin{cases} x+y-3=0,\\ x-y-1=0. \end{cases}$
解得$\begin{cases} x=2,\\ y=1. \end{cases}$
$\therefore \dfrac{1}{2}[(-x^{-2}y)^{2}]^{-3}$
$=\dfrac{1}{2}(-x^{-2}y)^{-6}$
$=\dfrac{1}{2}(x^{12} · y^{-6})$
$=\dfrac{1}{2}×2^{12}×1^{-6}$
$=2^{11}.$
6. 已知$a^2 - 4a + 1 = 0$，求$(a^2 - a^{-2})(a - a^{-1})$的值。
答案
解 $(a^{2}-a^{-2})(a-a^{-1})$
$=(a^{2}-\dfrac{1}{a^{2}})(a-\dfrac{1}{a})$
$=(a+\dfrac{1}{a})(a-\dfrac{1}{a})^{2}$
$=(a+\dfrac{1}{a})[(a+\dfrac{1}{a})^{2}-4].$
$\because a^{2}-4a+1=0,$
$\therefore a+\dfrac{1}{a}=4,$
$\therefore 原式=4×(4^{2}-4)=48.$
$=(a^{2}-\dfrac{1}{a^{2}})(a-\dfrac{1}{a})$
$=(a+\dfrac{1}{a})(a-\dfrac{1}{a})^{2}$
$=(a+\dfrac{1}{a})[(a+\dfrac{1}{a})^{2}-4].$
$\because a^{2}-4a+1=0,$
$\therefore a+\dfrac{1}{a}=4,$
$\therefore 原式=4×(4^{2}-4)=48.$
