数学分时根式化简

因为a/(b+c+d)=b/(a+c+d)
所以{a/(b+c+d)} +1 = {b/(a+c+d)} +1
所以{a/(b+c+d)} + {(b+c+d)/(b+c+d)} = {b/(a+c+d)} + {(a+c+d)/(a+c+d)}
所以 (a+b+c+d)/(b+c+d) = (a+b+c+d) / (a+c+d)
所以 (b+c+d)*(a+b+c+d) = (a+b+c+d)*(a+c+d) 分子分母交叉相乘
当(a+b+c+d) 不等于 0 时
所以 (b+c+d) = (a+c+d)
所以 a=b
类似可得 a=b=c=d
所以n= 1/3
当(a+b+c+d) 等于 0 时 无解 ----舍去

(√6+4√3+3√2)/[(√6+√3)(√3+√2)]
=[(√6+√3)+3(√3+√2)]/[(√6+√3)(√3+√2)]
=(√6+√3)/[(√6+√3)(√3+√2)]+3(√3+√2)/[(√6+√3)(√3+√2)]
=1/(√3+√2)+3/(√6+√3)
=√3-√2+√6-√3
=√6-√2

解：
xy/x-y=-1/3取倒数
x-y/xy=-3
1/x-1/y=3

分子和分母同时除以xy
(2x+3xy-2y)/(x-2xy-y)
=(2/y+3-2/x)/(1/y-2-1/x)
=(-6+3)/(-3-2)=3/5

1／a-1／b=1,
(b-a)/ab=1
b-a=ab
a-b=-ab

2a+3ab-2b／a-2ab-b
=(-2ab+3ab)/(-ab-2ab)
=ab/-3ab
=-1/3

1.因为a/(b+c+d)=b/(a+c+d)
所以{a/(b+c+d)} +1 = {b/(a+c+d)} +1
所以{a/(b+c+d)} + {(b+c+d)/(b+c+d)} = {b