试题

题目:
已知x2-xy=60,xy-y2=40,求代数式x2-y2和x2-2xy+y2的值.
答案
解:∵x2-xy=6u,xy-y2=它u,
∴x2-y2=(x2-xy)+(xy-y2)=1uu.
x2-2xy+y2=(x2-xy)-(xy-y2)=2u.
解:∵x2-xy=6u,xy-y2=它u,
∴x2-y2=(x2-xy)+(xy-y2)=1uu.
x2-2xy+y2=(x2-xy)-(xy-y2)=2u.
考点梳理
整式的加减.
将已知两等式相加即可求出x2-y2的值,两等式相减即可求出x2-2xy+y2的值.
此题考查了整式的加减,熟练掌握运算法则是解本题的关键.
计算题.
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