已知x=（根号2+根号3）⼀（根号2-根号3）求···

x=(√2+√3)/(√2-√3) = (√2+√3)²/(2-3) = -(√2+√3)²=-(5+2√6)
x²=[-(5+2√6)]²=49+20√6

(x²+10x+1)²+x²+20x+2
= [(49+20√6)-10(5-2√6)+1]² + 49+20√6-20(5+2√6)+2
= 0² - 51 - 20√6 +2
= -49 - 20√6

=============================================================================或者：
x=(√2+√3)/(√2-√3) = (√2+√3)²/(2-3) = -(√2+√3)²=-(5+2√6)
x²=[-(5+2√6)]²=49+20√6 = -1+10(5+2√6) = -1-10x

(x²+10x+1)²+x²+20x+2
= (-1-10x+10x+1)² -1-10x+20x+2
= 0²+10x+1
= -10(5+2√6)+1
= -49-20√6

x=(√2+√3)/(√2-√3) = (√2+√3)²/(2-3) = -(√2+√3)²=-(5+2√6)
∴x+5=-2√6
两边平方得：x²+10x+25=24
∴x²+10x=-1
(x²+10x+1)²+x²+20x+2
=(-1+1)²+-1+10x+2=10x+1
= -10(5+2√6)+1
= -49-20√6

先化简 X 再代入式子里面算