题目内容
(1)已知x+y=-4,xy=-12,求
+
的值;
(2)已知x2-3x+1=0,求x2-
的值.
y+1 |
x+1 |
x+1 |
y+1 |
(2)已知x2-3x+1=0,求x2-
1 |
x2 |
分析:(1)先根据分式混合运算的法则把原式进行化简,再把x+y=-4,xy=-12代入进行计算即可;
(2)先根据x2-3x+1=0求出x+
与x-
的值,再代入代数式进行计算即可.
(2)先根据x2-3x+1=0求出x+
1 |
x |
1 |
x |
解答:解:(1)原式=
=
=
,
∵x+y=-4,xy=-12,
∴原式=
=-
;
(2)∵x2-3x+1=0,
∴x≠0,
∴x-3+
=0,
∴x+
=3,
∴(x+
)2=9,
∴x2+
=7,
∵(x-
)2=x2+
-2=7-2=5,
∴x-
=±
,
∴x2-
=(x+
)(x-
)=±3
.
(y+1)2+(x+1)2 |
(x+1)(y+1) |
=
y2+2y+1++2x+1 |
xy+(x+y)+1 |
=
(x+y)2-2xy+2(x+y)+2 |
xy+(x+y)+1 |
∵x+y=-4,xy=-12,
∴原式=
(-4)2-2(-12)+2(-4)+2 |
-12+(-4)+1 |
=-
34 |
15 |
(2)∵x2-3x+1=0,
∴x≠0,
∴x-3+
1 |
x |
∴x+
1 |
x |
∴(x+
1 |
x |
∴x2+
1 |
x2 |
∵(x-
1 |
x |
1 |
x2 |
∴x-
1 |
x |
5 |
∴x2-
1 |
x2 |
1 |
x |
1 |
x |
5 |
点评:本题考查的是分式的化简求值,熟知分式混合运算的法则是解答此题的关键.
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