《2025年零障碍导教导学案八年级数学上册人教版》

(1)$10^{2}×10^{3}= (10×10)×(10×10×10)= $______;
(2)$a^{3}\cdot a^{4}= (a\cdot a\cdot a)\cdot (a\cdot a\cdot a\cdot a)= $______.
同底数幂的乘法法则:
同底数幂相乘,底数______,指数______,即$a^{m}\cdot a^{n}= a^{( )}$($m,n$都是正整数).

答案：
(1)$10^{5}$
(2)$a^{7}$
同底数幂的乘法法则:
不变 相加 $m+n$

1. 例(新教材P98探究改编)计算:
(1)$5^{2}×5^{4}= $______;
(2)$a\cdot a^{3}= $______;
(3)$x^{4}\cdot x^{8}= $______;
(4)$(\frac {1}{2})^{3}×(\frac {1}{2})^{2}= $______.

答案： 1.
(1)$5^{6}$
(2)$a^{4}$
(3)$x^{12}$
(4)$\left(\dfrac{1}{2}\right)^{5}$

2. (新教材P99 T2改编)计算:
(1)$3^{2}×3^{6}= $______;(2)$a^{4}\cdot a^{2}= $______;
(3)$x\cdot x^{2}= $______;(4)$a^{n}\cdot a^{2}= $______;
(5)$(-a)^{2}\cdot (-a)^{4}= $______;
(6)$(x-y)^{2}\cdot (x-y)^{3}= $______.

答案： 2.
(1)$3^{8}$
(2)$a^{6}$
(3)$x^{3}$
(4)$a^{2+n}$
(5)$a^{6}$
(6)$(x-y)^{5}$

3. (新教材P101习题T1改编)计算:
(1)$a^{2}\cdot a^{5}\cdot a^{8}= $______;
(2)$10^{4}×10×10^{2}= $______;
(3)$(\frac {1}{5})^{2}×(\frac {1}{5})^{3}×(\frac {1}{5})^{5}= $______;
(4)$(a+b)\cdot (a+b)^{3}\cdot (a+b)^{4}= $______.

答案： 3.
(1)$a^{15}$
(2)$10^{7}$
(3)$\left(\dfrac{1}{5}\right)^{10}$
(4)$(a+b)^{8}$

4. 例计算:$a\cdot a^{3}\cdot a^{5}-a^{2}\cdot a^{3}\cdot a^{4}$.

答案： 4.解:原式$=a^{1+3+5}-a^{2+3+4}$
$=a^{9}-a^{9}=0$.

5. 计算:$2x^{2}\cdot x^{4}\cdot x^{5}-x\cdot x^{4}\cdot x^{6}$.

答案： 5.解:原式$=2x^{2+4+5}-x^{1+4+6}$
$=2x^{11}-x^{11}=x^{11}$.

6. 例若$a^{m-1}\cdot a^{2}= a^{7}$,求$m$的值.

答案： 6.解:$\because a^{m-1}\cdot a^{2}=a^{7}$,
$\therefore a^{m-1+2}=a^{7}$.
$\therefore a^{m+1}=a^{7}$.
$\therefore m+1=7.\therefore m=6$.

7. 若$3^{n}×81= 3^{9}$,求$n$的值.

答案： 7.解:$\because 3^{n}× 81=3^{9}$,
$\therefore 3^{n}× 3^{4}=3^{9}$.
$\therefore 3^{n+4}=3^{9}$.
$\therefore n+4=9.\therefore n=5$.

8. 例若$x^{m}= 5,x^{n}= 6$,求$x^{m+n}$的值.

答案： 8.解:$\because x^{m}=5,x^{n}=6$,
$\therefore x^{m+n}=x^{m}\cdot x^{n}=5× 6=30$.

9. (2024·广州校级期中)若$3^{m}= 7,3^{n}= 6$,求$3^{m+n}$的值.

答案： 9.解:$\because 3^{m}=7,3^{n}=6$,
$\therefore 3^{m+n}=3^{m}\cdot 3^{n}=7× 6=42$.