已知x+3⼀x+2=1⼀ 根号2+1 求x-3⼀2x-4÷(5⼀x-2-x-2)的值

(x+3)/(x+2)=1/(√2+1)
x+2=(x+3)(√2+1)
x+2=(x+2+1)(√2+1)
x+2=(√2+1)(x+2)+√2+1
(x+2)√2=√2+1
x+2=(√2+1)/√2=(2+√2)/2
x+3=(4+√2)/2

原式=(x-3)/(2x-4)÷[5/(x-2)-x-2)
=(x-3)/(2x-4)÷(9-x^2)/(x-2)
=(x-3)/[2(x-2)]*[(x-2)/(9-x^2)]
=(x-3)/[2(x-2)]*(x-2)/[(3-x)(x+3)]
=-1/[2(x+3)]
=-1/[2(4+√2)/2]
=(√2-4)/12

忍不住问一句,你的题目有没有写错?

x+3/x+2=1/ 根号2+1
x+2=(x+3)(√2+1)
x+2=(x+2+1)(√2+1)
x+2=(√2+1)(x+2)+√2+1
(x+2)√2=√2+1
x+2=(√2+1)/√2=(2+√2)/2
x+3=(4+√2)/2
x-3/2x-4÷(5/x-2-x-2)==(x-3)/(2x-4)÷[5/(x-2)-x-2)
=(x-3)/(2x-4)÷(9-x^2)/(x-2)
=(x-3)/[2(x-2)]*[(x-2)/(9-x^2)]
=(x-3)/[2(x-2)]*(x-2)/[(3-x)(x+3)]
=-1/[2(x+3)]
=-1/[2(4+√2)/2]
=(√2-4)/12

计算这么烦