问题 解答题
若方程组
x2+y2=34
x-y=6
的解为
x=x1
y=y1
x=x2
y=y2

求:(1)
x21
+
x22

(2)
1
y1
+
1
y2
答案

x2+y2=34①
x-y=6②

由②得y=x-6③,

把③代入①得x2+(x-6)2=34,

整理得x2-6x+1=0,

根据题意得x1+x2=6,x1•x2=1,

(1)

x21
+
x22

=(x1+x22-2x1•x2

=36-2

=34;

(2)∵y=x-6,

∴y1=x1-6,y2=x2-6,

∴y1+y2=x1+x2-12=6-12=-6,

y1,•y2=x1•x2-6(x1+x2)+36=1-6×6+36=1,

1
y1
+
1
y2
=
y1+y2
y1y2
=
-6
1
=-6.

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