题目内容
10、若a2+2ab=-10,b2+2ab=16,则多项式a2+4ab+b2与a2-b2的值分别为( )
分析:将多项式合理变形即可,a2+4ab+b2=(a2+2ab)+(b2+2ab);a2-b2=(a2+2ab)-(b2+2ab).
解答:解:∵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.
∴a2+4ab+b2
=(a2+2ab)+(b2+2ab),
=-10+16,
=6;
∴a2-b2
=(a2+2ab)-(b2+2ab),
=-10-16,
=-26.
故选C.
点评: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.
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