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典例1
答案D
解析 $a=-0.2, b=-2^{-1}=-\dfrac{1}{2}=-0.5, c=(-\dfrac{1}{2})^{-2}=4$,
$\because -0.5＜-0.2＜4, \therefore b＜a＜c$, 故选 D.
典例2
答案D
解析 $a=-0.2, b=-2^{-1}=-\dfrac{1}{2}=-0.5, c=(-\dfrac{1}{2})^{-2}=4$,
$\because -0.5＜-0.2＜4, \therefore b＜a＜c$, 故选 D.
典例2
答案
解 (1) $(3x^2y^2z^{-1})^{-2} · (5xy^{-2}z^3)^2 =$
$( \dfrac{3x^2y^2}{z} )^{-2} · ( \dfrac{5xz^3}{y^2} )^2 = \dfrac{z^2}{9x^4y^4} · \dfrac{25x^2z^6}{y^4} = \dfrac{25z^8}{9x^2y^8}.$
(2) $(4x^2yz^{-1})^2 · (2xyz)^{-4} ÷ (yz^3)^{-2}$
$= \dfrac{16x^4y^2}{z^2} · \dfrac{1}{16x^4y^4z^4} ÷ \dfrac{1}{y^2z^6}$
$= \dfrac{1}{z^6y^2} · y^2z^6 = 1.$
$( \dfrac{3x^2y^2}{z} )^{-2} · ( \dfrac{5xz^3}{y^2} )^2 = \dfrac{z^2}{9x^4y^4} · \dfrac{25x^2z^6}{y^4} = \dfrac{25z^8}{9x^2y^8}.$
(2) $(4x^2yz^{-1})^2 · (2xyz)^{-4} ÷ (yz^3)^{-2}$
$= \dfrac{16x^4y^2}{z^2} · \dfrac{1}{16x^4y^4z^4} ÷ \dfrac{1}{y^2z^6}$
$= \dfrac{1}{z^6y^2} · y^2z^6 = 1.$
答案
解：
(1)
$\begin{aligned}(3x^2y^2z^{-1})^{-2} · (5xy^{-2}z^3)^2&= 3^{-2}x^{-4}y^{-4}z^{2} · 5^2x^2y^{-4}z^6 \\&= \frac{1}{9}x^{-4}y^{-4}z^2 · 25x^2y^{-4}z^6 \\&= \frac{25z^8}{9x^2y^8}\end{aligned}$
(2)
$\begin{aligned}(4x^2yz^{-1})^2 · (2xyz)^{-4} ÷ (yz^3)^{-2}&= \frac{16x^4y^2}{z^2} · \frac{1}{16x^4y^4z^4} ÷ \frac{1}{y^2z^6} \\&= \frac{1}{y^2z^6} · y^2z^6 \\&= 1\end{aligned}$
(1)
$\begin{aligned}(3x^2y^2z^{-1})^{-2} · (5xy^{-2}z^3)^2&= 3^{-2}x^{-4}y^{-4}z^{2} · 5^2x^2y^{-4}z^6 \\&= \frac{1}{9}x^{-4}y^{-4}z^2 · 25x^2y^{-4}z^6 \\&= \frac{25z^8}{9x^2y^8}\end{aligned}$
(2)
$\begin{aligned}(4x^2yz^{-1})^2 · (2xyz)^{-4} ÷ (yz^3)^{-2}&= \frac{16x^4y^2}{z^2} · \frac{1}{16x^4y^4z^4} ÷ \frac{1}{y^2z^6} \\&= \frac{1}{y^2z^6} · y^2z^6 \\&= 1\end{aligned}$
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
1. C
2.计算：$(a^{-1}b^{2})^{3}=$
$\frac{b^{6}}{a^{3}}$
.答案
2. $\frac{b^{6}}{a^{3}}$
3.若$ a^{-1}=(-1)^0 $，则$ a= $
1
.答案
3. 1
4.计算:
(1)$(2m^2n^{-3})^2 · (-mn^{-2})^{-2}$；
(2)$4x^2y^{-3}z ÷ (-2x^{-1}yz^{-2})^2$；
(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)$(2m^2n^{-3})^2 · (-mn^{-2})^{-2}$；
(2)$4x^2y^{-3}z ÷ (-2x^{-1}yz^{-2})^2$；
(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
4. 解 (1)原式=$4m^4n^{-6} · m^{-2}n^4$
=$4m^2n^{-2}$
=$\frac{4m^2}{n^2}.$
(2)原式=$4x^2y^{-3}z ÷ (4x^{-2}y^2z^{-4})$
=$x^4y^{-5}z^5$
=$\frac{x^4z^5}{y^5}.$
(3)原式=$8−8×0.125+1+1$
=$8−1+1+1$
=$9.$
(4)原式=$2×1+8×\frac{1}{6}+16$
=$2+\frac{4}{3}+16=19\frac{1}{3}.$
=$4m^2n^{-2}$
=$\frac{4m^2}{n^2}.$
(2)原式=$4x^2y^{-3}z ÷ (4x^{-2}y^2z^{-4})$
=$x^4y^{-5}z^5$
=$\frac{x^4z^5}{y^5}.$
(3)原式=$8−8×0.125+1+1$
=$8−1+1+1$
=$9.$
(4)原式=$2×1+8×\frac{1}{6}+16$
=$2+\frac{4}{3}+16=19\frac{1}{3}.$
5.已知$|x + y - 3| + (x - y - 1)^2 = 0$，求$\frac{1}{2}[(-x^{-2}y)^2]^{-3}$的值。
答案
5. 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$\begin{cases}x=2,\\y=1.\end{cases}$
$\therefore \frac{1}{2}[(-x^{-2}y)^2]^{-3}$
=$\frac{1}{2}(-x^{-2}y)^{-6}$
=$\frac{1}{2}(x^{12} · y^{-6})$
=$\frac{1}{2}×2^{12}×1^{-6}$
=$2^{11}.$
$\therefore \begin{cases}x+y-3=0,\\x-y-1=0.\end{cases}$
解得$\begin{cases}x=2,\\y=1.\end{cases}$
$\therefore \frac{1}{2}[(-x^{-2}y)^2]^{-3}$
=$\frac{1}{2}(-x^{-2}y)^{-6}$
=$\frac{1}{2}(x^{12} · y^{-6})$
=$\frac{1}{2}×2^{12}×1^{-6}$
=$2^{11}.$
6. 已知$a^2 - 4a + 1 = 0$，求$(a^2 - a^{-2})(a - a^{-1})$的值。
答案
6. 解 $(a^2-a^{-2})(a-a^{-1})$
=$(a^2-\frac{1}{a^2})(a-\frac{1}{a})$
=$(a+\frac{1}{a})(a-\frac{1}{a})^2$
=$(a+\frac{1}{a})[(a+\frac{1}{a})^2-4].$
$\because a^2-4a+1=0,\therefore a+\frac{1}{a}=4,$
$\therefore 原式=4×(4^2-4)=48.$
=$(a^2-\frac{1}{a^2})(a-\frac{1}{a})$
=$(a+\frac{1}{a})(a-\frac{1}{a})^2$
=$(a+\frac{1}{a})[(a+\frac{1}{a})^2-4].$
$\because a^2-4a+1=0,\therefore a+\frac{1}{a}=4,$
$\therefore 原式=4×(4^2-4)=48.$
