Question

化简求值[(x^2-16/x^2+8x+16)+(x/x-4)]÷1/x^2-16，其中x=√2+1（不是根号下2+1，也就是不是跟3）

Answer 1

x=√2+1
x-1=√2
两边平方
x²-2x+1=2
x²-2x=1

原式=[(x+4)(x-4)/(x+4)²+x/(x-4)]×(x²-16)
=[(x-4)/(x+4)+x/(x-4)]×(x+4)(x-4)
=(x-4)/(x+4)×(x+4)(x-4)+x/(x-4)×(x+4)(x-4)

=(x-4)²+x(x+4)
=x²-8x+16+x²+4x
=2x²-4x+16
=2(x²-2x)+16
=2+16
=18

Answer 2

原式=((X+4)^2-16/X^2))+(X/X-4)
(X^2-16)
=(X+4+4/X)(X+4-4/X)+X(X
+4)
=(√2+1+4+4/(√2+1)
(√2+1+4-4/(√2+1)+(√2+1)
(√2+1+4)
=(√2+5+4√2-4)
(√2+5-4√2+4)+(√2+1)(√2+5)
=(5√2-1)(-3√2+9)+(2+6√2+5)
=-30+45√2+3√2-9+2+6√2+5
=-32+54√2