若a2+2ab=-10,b2+2ab=16,则多项式a2+4ab+b2与a2-b2的值分别为(  )A. 6,26B. -6,26C. 6,-26D. -6,-26

问题描述:

若a2+2ab=-10,b2+2ab=16,则多项式a2+4ab+b2与a2-b2的值分别为(  )
A. 6,26
B. -6,26
C. 6,-26
D. -6,-26

∵a2+2ab=-10,b2+2ab=16,
∴a2+4ab+b2
=(a2+2ab)+(b2+2ab),
=-10+16,
=6;
∴a2-b2
=(a2+2ab)-(b2+2ab),
=-10-16,
=-26.
故选C.
答案解析:将多项式合理变形即可,a2+4ab+b2=(a2+2ab)+(b2+2ab);a2-b2=(a2+2ab)-(b2+2ab).
考试点:整式的加减—化简求值.
知识点:解答本题的关键是合理的将多项式进行变形,与已知相结合.