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计算:(1)(sin30°)?2+(π?2)0?|3?18|+(?1)2011;(2)先化简,再求值:(x?1x-x?2x+1)÷2x2?xx2+2x+1
计算:(1)(sin30°)?2+(π?2)0?|3?18|+(?1)2011;(2)先化简,再求值:(x?1x-x?2x+1)÷2x2?xx2+2x+1,其中x满足x2...
计算:(1)(sin30°)?2+(π?2)0?|3?18|+(?1)2011;(2)先化简,再求值:(x?1x-x?2x+1)÷2x2?xx2+2x+1,其中x满足x2-x-1=0.
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(1)原式=(
)-2+1-3
+3-1
=4+1-3
+3-1
=7-3
;
(2)∵x2-x-1=0,
∴x2=x+1,
原式=
×
=
×
=
,
当x2=x+1时,原式=
=1.
1 |
2 |
2 |
=4+1-3
2 |
=7-3
2 |
(2)∵x2-x-1=0,
∴x2=x+1,
原式=
(x+1)(x?1)?x(x?2) |
x(x+1) |
(x+1)2 |
x(2x?1) |
=
2x?1 |
x(x+1) |
(x+1)2 |
x(2x?1) |
=
x+1 |
x2 |
当x2=x+1时,原式=
x+1 |
x+1 |
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