例已知a=51+20,b=p-30=2√2 q=3,a-b=√593,求a+b 解:a+b=6p-q a+b=(a+b)+b)=(6p0)6p =36p·+@q-12pq=36+q2-12pq 36×8+9-12pq 又∵-b=4p+5 a-b2=(a-b)(a-b)=(4p+50)(4+0) =16·p+254·q+40D·q 16×8+25×9+40D·q=593→pf=6 则a a+b=153, 593, . 5 2 , 3 , 2 2, q a b a b a p q b p q p             = − = + = + = − = 求 例 已知 p p q q p q a b a b a b p q p q                 =  +  −   + = +  + = −  − 3 6 1 2 ( ) ( ) (6 ) (6 ) 2 p q p q     = 36 + − 12  2 2 a b p q     解  + = 6 − p p q q p q a b a b a b p q p q                 =  +  +   − = −  − = +  + 1 6 2 5 4 0 ( ) ( ) (4 5 ) (4 5 ) 2 a b p q     又 − = 4 + 5 = 593 p q   = 368 + 9 − 12  p q   = 168 + 259 + 40   p  q = 6   a + b = 15.   则