发布网友 发布时间:2024-10-02 00:29
共4个回答
热心网友 时间:2024-11-06 02:11
一式:(a+b)^2=a^2+2ab+b^2=7,二式:(a-b)^2=a^2-2ab+b^2=3,一式减二式:4ab=4.ab=1;a^2+b^2=(a+b)^2-2ab=7*2-2*1=12热心网友 时间:2024-11-06 02:18
ab的值是1a2+b2的值是5热心网友 时间:2024-11-06 02:11
这题就是公式的变形的运用,挺巧的解:ab=[(a+b)�0�5-(a-b)�0�5]/4=(7�0�5-3�0�5)/4=10a�0�5+b�0�5=[(a+b)�0�5+(a-b)�0�5]/2=(7+3)/2=5 仅供参考热心网友 时间:2024-11-06 02:13
(a+b)�0�5=7,(a-b)�0�5=3,a�0�5+b�0�5+2ab=7,a�0�5+b�0�5-2ab=3a�0�5+b�0�5=5ab=1热心网友 时间:2024-11-06 02:11
一式:(a+b)^2=a^2+2ab+b^2=7,二式:(a-b)^2=a^2-2ab+b^2=3,一式减二式:4ab=4.ab=1;a^2+b^2=(a+b)^2-2ab=7*2-2*1=12热心网友 时间:2024-11-06 02:12
这题就是公式的变形的运用,挺巧的解:ab=[(a+b)�0�5-(a-b)�0�5]/4=(7�0�5-3�0�5)/4=10a�0�5+b�0�5=[(a+b)�0�5+(a-b)�0�5]/2=(7+3)/2=5 仅供参考热心网友 时间:2024-11-06 02:11
(a+b)�0�5=7,(a-b)�0�5=3,a�0�5+b�0�5+2ab=7,a�0�5+b�0�5-2ab=3a�0�5+b�0�5=5ab=1热心网友 时间:2024-11-06 02:17
ab的值是1a2+b2的值是5