已知(x+y)2=49,(x﹣y)2=1,求下列各式的值:
(1)x2+y2;(2)xy.
解:由题意知:(x+y)2=x2+y2+2xy=49①,
(x﹣y)2=x2+y2﹣2xy=1②,
①+②得:(x+y)2+(x﹣y)2,
=x2+y2+2xy+x2+y2﹣2xy,
=2(x
化简1x+1+12x+2= 32x+2 ,当x=____ 时,原式的值等于1.
— Is this your mother?
— _______. She's my aunt.
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