《2025年培优课堂七年级数学上册华师大版四川专版》

$3. $观察下列式子$,$并回答问题：　　
$ 1^{3}= \frac{1}{4}×1^{2}×2^{2} ；$$ 1^{3}+2^{3}= \frac{1}{4}×2^{2}×3^{2} ；$　　
$ 1^{3}+2^{3}+3^{3}= \frac{1}{4}×3^{2}×4^{2} ；$$ 1^{3}+2^{3}+3^{3}+4^{3}= \frac{1}{4}×4^{2}×5^{2} ；$　　
$…$　　
计算：$ 7^{3}+8^{3}+9^{3}+…+19^{3} .$　　

答案： 解:原式$=1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}+9^{3}+\cdots +19^{3}-(1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3})=\dfrac{1}{4}× 19^{2}× 20^{2}-\dfrac{1}{4}× 6^{2}× 7^{2}=\dfrac{1}{4}× 361× 400-\dfrac{1}{4}× 36× 49=361× 100-9× 49=36100-441=35659$.

$4. $计算：$ 2023×(\frac{1}{1011}-\frac{1}{1012})+1011×(\frac{1}{1012}+\frac{1}{2023})-1012×(\frac{1}{1011}-\frac{1}{2023}) .$　　

答案： 解:原式$=(1011+1012)× \left(\dfrac{1}{1011}-\dfrac{1}{1012}\right)+1011× \left(\dfrac{1}{1012}+\dfrac{1}{2023}\right)-1012× \left(\dfrac{1}{1011}-\dfrac{1}{2023}\right)=1011× \left(\dfrac{1}{1011}-\dfrac{1}{1012}\right)+1012× \left(\dfrac{1}{1011}-\dfrac{1}{1012}\right)+1011× \left(\dfrac{1}{1012}+\dfrac{1}{2023}\right)-1012× \left(\dfrac{1}{1011}-\dfrac{1}{2023}\right)=1-\dfrac{1011}{1012}+\dfrac{1012}{1011}-1+\dfrac{1011}{1012}+\dfrac{1011}{2023}-\dfrac{1012}{1011}+\dfrac{1012}{2023}=\dfrac{1011}{2023}+\dfrac{1012}{2023}=\dfrac{2023}{2023}=1$.