Question

（1）已知a+b=2。，ab=1，求a²+b²，（a-b）²的值；（2）若a+1/a=3，求a²+1/a²的

Answer 1

解：
1.a²+b²=（a+b）²-2ab=a²+2ab+b²-2ab=2²-2×1=2
（a-b）²=a²-2ab+b²=2-2ab=2-2×1=0
2.a+1/a=3
（a+1/a）²=9
a²+2×a×1/a+1/a²=9
a²+1/a²+2=9
a²+1/a²=7

Answer 2

a²+b²=a²+b²+2ab-2ab=（a+b）²-2ab=2²-2*1=2
a²+1/a²=a²+1/a²+2*a*1/a-2*a*1/a=(a+1/a)²-2*a*1/a=3²-2=7

Answer 3

a^2+b^2=2
(a-b)^2=0
（2）7

Answer 4

a²+b²=（a+b）的平方-2ab=2

（a-b）²=a²+b²-2ab=0

a+1/a=3，a²+1/a²=（a+1/a）的平方-2=7

Answer 5

解(1)已知a+b=2。，ab=1
所以a²+b²=（a+b）^2-2ab=4-2×1=2
（a-b）^2=(a+b）^2-4ab=4-4×1=0
(2)若a+1/a=3
则a^2+1/a^2=（a+1/a）^2-2=9-2=7

Answer 6

a²+b²=（a+b）²-2ab=4-2=2;
(a-b)²=a²-2ab+b²=(a²+b²)-2ab=2-2=0;
a²+1/a²=(a+1/a)-2=3-2=1