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(1)已知x+y=-4,xy=-10,求y+1x+1+x+1y+1的值.(2)已知x2+3x-1=0,求x4−2x2+1x2和x2x4+1的值.

题目详情
(1)已知x+y=-4,xy=-10,求
y+1
x+1
+
x+1
y+1
的值.
(2)已知x2+3x-1=0,求
x4−2x2+1
x2
x2
x4+1
的值.
▼优质解答
答案和解析
(1)原式=
(y+1)2+(x+1)2
(x+1)(y+1)

=
(x2+y2)+2(x+y)+2
xy+(x+y)+1

=
(x+y)2−2xy+2(x+y)+2
xy+(x+y)+1

∵x+y=-4,xy=-10,
∴原式=
(−4)2−2×(−10)+2×(−4)+2
−10+(−4)+1

=
16+20−8+2
−13

=-
30
13


(2)∵x2+3x-1=0,
∴x2=1-3x,
原式=
(x2−1)2
x2

=
(1−3x−1)2
x2

=
9x2
x2

=9;
原式=
x2
(1−3x)2+1

=
x2
9x2−6x+2

=
1−3x
11(1−3x)

=
1
11