Question

已知x+3/x+2=1/(√3+√2+1),求(x-3/2x-4)/((5/x-2)-x-2)

要求写详细过程

Answer 1

前一步与一楼一样
(x+2)/(x+3)=√3+√2+1
(x+2)/(x+3)-1=√3+√2+1-1
(x+2-x-3)/(x+3)=√3+√2
-1/(x+3)=√3+√2

第二步 (x-3/2x-4)/((5/x-2)-x-2)
=(x-3)/(2x-4)÷【5/(x-2)-｛（x+2)(x-2)}/(x-2)】
=(x-3)(2x-4)÷(5-x的二次方+4)/(x-2）
=（x-3）/(2x-4)*(x-2)/(9-x的平方）
=-1/(6+6x)
把-1/(x+3)=√3+√2带入得(√3+√2)/2

Answer 2

取倒数
(x+2)/(x+3)=√3+√2+1
(x+2)/(x+3)-1=√3+√2+1-1
(x+2-x-3)/(x+3)=√3+√2
-1/(x+3)=√3+√2

原式=[(x-3)/2(x-2)]/[(5-x²+4)/(x-2)]
=[(x-3)/2(x-2)]/[-(x+3)(x-3)/(x-2)]
=-1/[2(x+3)]
=[-1/(x+3)]/2
=(√3+√2)/2