高一数学若f(x)=ax+6/x-3,且f(根号2+1)=0,求f(1/1-根号2)的值根号下只包含2

Question

高一数学若f(x)=ax+6/x-3,且f(根号2+1)=0,求f(1/1-根号2)的值
根号下只包含2

Answer 1

令,1/(1-√2)=t,有√2+1=-t,t=-(√2+1).f(x)=(ax+6)/(x-3),∵f(√2+1)=0,∴f(√2+1)=[a(√2+1)+6]/[(√2+1)-3]=[a(√2+1)-6]/(√2+4)=0,a(√2+1)-6=0,a=6(√2-1).f(x)=[6(√2-1)x+6]/(x-3).f[1/(1-√2)]=f[-(√2+1)]...