已知a大于0 且a不等于1,f(x)=1/(a^x+根号a)求值：f(0)+f(1),f(-1)+f(2)

由1的结果归纳概括对所有实数x都成立的一个等式，并加以证明

f(0) = 1/(a^0 + √a) = 1/(1 + √a) = (√a - 1)/[(√a + 1)(√a - 1)] = (√a - 1)/(a - 1)
f(1) = 1/(a^1+ √a) = 1/(a + √a) = (a - √a)/[(√a + a)(a - √a)] = (a - √a)/(a² - a) = (a - √a)/a(a - 1)
f(0) + f(1) = (√a - 1)/(a - 1) + (a - √a)/a(a - 1) = √a/a

f(-1) = 1/[a^(-1) + √a] = a/(1 + a√a) = a(a√a - 1)/[(a√a + 1)(a√a - 1)] = a(a√a - 1)/(a³ - 1)
f(2) = 1/(a² + √a) = (a² - √a)/[(a² + √a)(a² - √a)] = (a² - √a)/a(a³ - 1)
f(-1) + f(2) = a(a√a - 1)/(a³ - 1) + (a² - √a)/a(a³ - 1) = √a/a

对于任意的x、y，满足x+y = 1,等式：
f(x) + f(y) = √a/a = 1/√a。

证明: ∵ x + y = 1;
∴y = 1 - x
f(x) = 1/(a^x + √a) = (a^x - √a)/a[a^(2x-1) - 1]
f(y) = f(1-x) = 1/[a^(1-x) + √a]
= a^x / a^x[a^(1-x) + √a]
= a^x / (a + a^x√a)
= a^x(a^x√a - a) / [(a + a^x√a)(a^x√a - a)]
= a^x(a^x√a - a) / a²[a^(2x-1) - 1]

f(x) + f(y) = (a^x - √a)/a[a^(2x-1) - 1] + a^x(a^x√a - a) / a²[a^(2x-1) - 1]
= [a^(x+1) - a√a + a^2x√a - a^(x+1)]/a²[a^(2x-1) - 1]
= a√a[a^(2x-1) - 1]/a²[a^(2x-1) - 1]
= √a/a

= 1/√a

化简分式，通分运算就得出结果。 f(x)+f(1-x) =-a^0.5/(a^x+a^0.5)-a^0.5/(a^(1-x)+a^0.5) =-a^0.5(a^(1-x)+a^0.5)+a^0.5(a^x+a^0.5))/((a^x+a^0.5)(a^(1-x)+a^0.5)) =-(a^(0.5+1-x)+a+a^(0.5+x)+a)/(a+a^(0.5+1-x)+a^(0.5+x)+a) =-(2a+a^(0.5+1-x)+a^(0.5+x))/(2a+a^(0.5+1-x)+a^(0.5+x)) =-1

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