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1. (教材尝试变式)代数式 $6^{3}×6^{3}×6^{3}×6^{3}×6^{3}$ 可表示为(
A.$5×6^{3}$
B.$6^{3 + 5}$
C.$(6^{3})^{5}$
D.$(5×6)^{3}$
C
)A.$5×6^{3}$
B.$6^{3 + 5}$
C.$(6^{3})^{5}$
D.$(5×6)^{3}$
答案
1. C 解析：$6^{3}×6^{3}×6^{3}×6^{3}×6^{3}=(6^{3})^{5}$.
2. 下列计算正确的是(
A.$x^{3}·x^{4}=x^{12}$
B.$(x^{3})^{4}=x^{7}$
C.$(-x^{3})^{4}=-x^{12}$
D.$(-x^{4})^{3}=-x^{12}$
D
)A.$x^{3}·x^{4}=x^{12}$
B.$(x^{3})^{4}=x^{7}$
C.$(-x^{3})^{4}=-x^{12}$
D.$(-x^{4})^{3}=-x^{12}$
答案
2. D 解析：$x^{3}· x^{4}=x^{3 + 4}=x^{7}$，故A选项不符合题意；$(x^{3})^{4}=x^{3×4}=x^{12}$，故B选项不符合题意；$(-x^{3})^{4}=x^{3×4}=x^{12}$，故C选项不符合题意；$(-x^{4})^{3}=-x^{4×3}=-x^{12}$，故D选项符合题意.
3. 若 $(3×3×3×3)^{m}=9^{2}$，则 $m$ 的值为(
A.1
B.2
C.3
D.4
A
)A.1
B.2
C.3
D.4
答案
3. A 解析：由题意可得，$(3^{4})^{m}=(3^{2})^{2}$，$\therefore 3^{4m}=3^{4}$，$\therefore 4m = 4$，$\therefore m = 1$.
4. $x^{3m + 1}$ 也可以写成(
A.$(x^{3})^{m + 1}$
B.$(x^{m})^{3}+1$
C.$x^{m^{3}}·x$
D.$(x^{m})^{3}·x$
D
)A.$(x^{3})^{m + 1}$
B.$(x^{m})^{3}+1$
C.$x^{m^{3}}·x$
D.$(x^{m})^{3}·x$
答案
4. D 解析：$x^{3m + 1}=x^{3m}· x=(x^{m})^{3}· x$.
$5. (1)① [(\frac{1}{3})^{2}]^{4}=$；$② (a^{5})^{6}=$；$③ [(-8)^{3}]^{2}=$.
$(2)① (x^{3})^{2n}=$；$② (a^{2})^{n}·a^{3}=$；$③ (a^{2})^{4}·(-a)^{3}=$.
$(3)① x^{12}=( ) )^{6}=( ) )^{4}=( ) )^{3}=( ) )^{2}$；$② (a^{2})^{( ) )}·a^{3}=a^{11}.$
$(2)① (x^{3})^{2n}=$；$② (a^{2})^{n}·a^{3}=$；$③ (a^{2})^{4}·(-a)^{3}=$.
$(3)① x^{12}=( ) )^{6}=( ) )^{4}=( ) )^{3}=( ) )^{2}$；$② (a^{2})^{( ) )}·a^{3}=a^{11}.$
答案
5. (1)①$\frac{1}{3^{8}}$ ②$a^{30}$ ③$8^{6}$ 解析：①$[(\frac{1}{3})^{2}]^{4}=(\frac{1}{3})^{8}=\frac{1}{3^{8}}$；②$(a^{5})^{6}=a^{5×6}=a^{30}$；③$[(-8)^{3}]^{2}=(-8)^{3×2}=(-8)^{6}=8^{6}$.
(2)①$x^{6n}$ ②$a^{2n + 3}$ ③$-a^{11}$ 解析：①$(x^{3})^{2n}=x^{3· 2n}=x^{6n}$；②$(a^{2})^{n}· a^{3}=a^{2n}· a^{3}=a^{2n + 3}$；③$(a^{2})^{4}· (-a)^{3}=a^{8}· (-a^{3})=-a^{8 + 3}=-a^{11}$.
(3)①$x^{2}$ $x^{3}$ $x^{4}$ $x^{5}$ ②4
(2)①$x^{6n}$ ②$a^{2n + 3}$ ③$-a^{11}$ 解析：①$(x^{3})^{2n}=x^{3· 2n}=x^{6n}$；②$(a^{2})^{n}· a^{3}=a^{2n}· a^{3}=a^{2n + 3}$；③$(a^{2})^{4}· (-a)^{3}=a^{8}· (-a^{3})=-a^{8 + 3}=-a^{11}$.
(3)①$x^{2}$ $x^{3}$ $x^{4}$ $x^{5}$ ②4
6. (1)若 $a^{2x}=3$，则 $a^{4x}=$
(2)若 $3x + 2y - 2 = 0$，则 $8^{x}·4^{y}=$
9
.(2)若 $3x + 2y - 2 = 0$，则 $8^{x}·4^{y}=$
4
.答案
6. (1)9 解析：$a^{4x}=(a^{2x})^{2}=3^{2}=9$.
(2)4 解析：$\because 3x + 2y - 2 = 0$，$\therefore 3x + 2y = 2$，$\therefore 8^{x}· 4^{y}=(2^{3})^{x}· (2^{2})^{y}=2^{3x + 2y}=2^{2}=4$.
(2)4 解析：$\because 3x + 2y - 2 = 0$，$\therefore 3x + 2y = 2$，$\therefore 8^{x}· 4^{y}=(2^{3})^{x}· (2^{2})^{y}=2^{3x + 2y}=2^{2}=4$.
7. 计算：
(1) $(8^{5})^{6}$；
(2) $[(-a)^{3}]^{5}$；
(3) $a^{3}·a^{5}·(-a^{2})^{4}$.
(1) $(8^{5})^{6}$；
(2) $[(-a)^{3}]^{5}$；
(3) $a^{3}·a^{5}·(-a^{2})^{4}$.
答案
7. (1)原式$=8^{6×6}=8^{30}$.
(2)原式$=(-a)^{3×5}=(-a)^{15}=-a^{15}$.
(3)原式$=a^{3}· a^{5}· a^{8}=a^{3 + 5 + 8}=a^{16}$.
(2)原式$=(-a)^{3×5}=(-a)^{15}=-a^{15}$.
(3)原式$=a^{3}· a^{5}· a^{8}=a^{3 + 5 + 8}=a^{16}$.
8. 计算：
(1) $(-a^{2})^{3}·(-a^{3})^{4}$；
(2) $(a^{2})^{3}+5a^{3}·a^{3}$；
(3) $2(x^{2})^{3}-x^{2}·x^{4}$；
(4) $2x^{4}·x^{2}+(x^{3})^{2}-5(x^{2})^{3}$.
(1) $(-a^{2})^{3}·(-a^{3})^{4}$；
(2) $(a^{2})^{3}+5a^{3}·a^{3}$；
(3) $2(x^{2})^{3}-x^{2}·x^{4}$；
(4) $2x^{4}·x^{2}+(x^{3})^{2}-5(x^{2})^{3}$.
答案
8. (1)原式$=-a^{6}· a^{12}=-a^{6 + 12}=-a^{18}$.
(2)原式$=a^{6}+5a^{6}=6a^{6}$.
(3)原式$=2x^{6}-x^{6}=x^{6}$.
(4)原式$=2x^{6}+x^{6}-5x^{6}=-2x^{6}$.
(2)原式$=a^{6}+5a^{6}=6a^{6}$.
(3)原式$=2x^{6}-x^{6}=x^{6}$.
(4)原式$=2x^{6}+x^{6}-5x^{6}=-2x^{6}$.
