(1)已知f(1/X+1)=X²+1/x²+3/X,求f(x) (2)已知2f(-x)+f(x)=x,求f(x)
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![(1)已知f(1/X+1)=X²+1/x²+3/X,求f(x) (2)已知2f(-x)+f(x)=x,求f(x)](/uploads/image/z/8664923-11-3.jpg?t=%EF%BC%881%EF%BC%89%E5%B7%B2%E7%9F%A5f%EF%BC%881%2FX%2B1%29%3DX%26%23178%3B%2B1%2Fx%26%23178%3B%2B3%2FX%2C%E6%B1%82f%EF%BC%88x%EF%BC%89+%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A52f%EF%BC%88-x%EF%BC%89%2Bf%EF%BC%88x%EF%BC%89%3Dx%2C%E6%B1%82f%28x%29)
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(1)已知f(1/X+1)=X²+1/x²+3/X,求f(x) (2)已知2f(-x)+f(x)=x,求f(x)
(1)已知f(1/X+1)=X²+1/x²+3/X,求f(x) (2)已知2f(-x)+f(x)=x,求f(x)
(1)已知f(1/X+1)=X²+1/x²+3/X,求f(x) (2)已知2f(-x)+f(x)=x,求f(x)
(1)令1/x+1=t,可得1/x=t-1
f(x)=1+1/x^2+3/x,两边把1/x换成t-1,
(2)等式两边把所有x换成-x得到另一等式,和条件组成方程组,消去f(-x)
另t=1/x,则f(t+1)=1+t^2+3t=(t+1)^2+(t+1)-1,所以f(x)=x^2+X-1
(1)令x=1/t 带入得f(1+t)=t2+3t+1,令t=x-1 即可求得f(x)=x2+x-1
(2)思路同上,f(x)+2f(-x)=x f(-x)+2f(x)=-x 得f(x)=-f(-x) 带入2f(-x)+f(x)=x 求得f(x)=-x