Question

已知X1、X2是方程3X^2-2X-2=0的两根，不解方程，利用前面的结论求下列各式的值：1，1/X1^2+1/x^2; 2题X1X2

Answer 1

1/(x1)²
+
1/(x2)²
=[(x1)²
+
(x2)²]/[(x1)²(x2)²]
=[(x1
+
x2)²
-
2x1x2]/[(x1)²(x2)²]
因为
x1+x2=-b/a=2/3
x1*x2=c/a
=
-2/3
所以：
[(x1
+
x2)²
-
2x1x2]/[(x1)²(x2)²]
=[(2/3)²
-
2×(-2/3)]
/
[(-2/3)²]
=(4/9
+
4/3)
/(4/9)
=(16/9)
/
(4/9)
=
4
X1X2²+X1²X2
=(x1x2)(x1
+
x2)
=(-2/3)×(2/3)
=-4/9
x1^2+x2^2
=x1²+2x1x2+x2²
-
2x1x2
=(x1+x2)²
-
2x1x2
=(2/3)²
-
2×(-2/3)
=
4/9
+
4/3
=
16/9
这个题主要考的是ax²+bx+c=0中：
x1+x2=-b/a=2/3
x1*x2=c/a
=
-2/3