高一函数解方程组法原理最详实的解释原题:如函数f(x)满足af(x)+f(1/x)=ax,x∈R且x≠0,a为常数,a≠±1,求f(x)?书上的解析是:af(x)+f(1/x)=ax 通过1/x换成x得到一个新的式子 af(1/x)+f(x)=a/x1.为什么换
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/08 05:25:31
![高一函数解方程组法原理最详实的解释原题:如函数f(x)满足af(x)+f(1/x)=ax,x∈R且x≠0,a为常数,a≠±1,求f(x)?书上的解析是:af(x)+f(1/x)=ax 通过1/x换成x得到一个新的式子 af(1/x)+f(x)=a/x1.为什么换](/uploads/image/z/3510738-18-8.jpg?t=%E9%AB%98%E4%B8%80%E5%87%BD%E6%95%B0%E8%A7%A3%E6%96%B9%E7%A8%8B%E7%BB%84%E6%B3%95%E5%8E%9F%E7%90%86%E6%9C%80%E8%AF%A6%E5%AE%9E%E7%9A%84%E8%A7%A3%E9%87%8A%E5%8E%9F%E9%A2%98%EF%BC%9A%E5%A6%82%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3af%28x%29%2Bf%281%2Fx%29%3Dax%2Cx%E2%88%88R%E4%B8%94x%E2%89%A00%EF%BC%8Ca%E4%B8%BA%E5%B8%B8%E6%95%B0%EF%BC%8Ca%E2%89%A0%C2%B11%EF%BC%8C%E6%B1%82f%28x%29%3F%E4%B9%A6%E4%B8%8A%E7%9A%84%E8%A7%A3%E6%9E%90%E6%98%AF%EF%BC%9Aaf%28x%29%2Bf%281%2Fx%29%3Dax+%E9%80%9A%E8%BF%871%2Fx%E6%8D%A2%E6%88%90x%E5%BE%97%E5%88%B0%E4%B8%80%E4%B8%AA%E6%96%B0%E7%9A%84%E5%BC%8F%E5%AD%90+af%281%2Fx%29%2Bf%28x%29%3Da%2Fx1.%E4%B8%BA%E4%BB%80%E4%B9%88%E6%8D%A2)
xTnF@)D.IRXCby6'M"*eע|Km-MЦe$%
.9sv;ɯI_:iDslG[ ?|yyEzоYExm&jQu?v_-?%@c'zȀ'u-ym%Vy<>?r:aZ2+-k FaA{a0"c)P^!e+<
2= `գ䥜SBȄcWM:CMMl78$@YYVA>jȅ&uX$znSw
&V0hF5,
a6ΧF0sC:2Ƀ&+Hי}{']iojD!Xl g9)&ǃf%[NAF*wf?
p`|nWřW_@<>03Wʗʕ\5mޘ}_*xtCL§*sEH6dɒx4lj%:,NƲh˂$4DQTBt( ކR
#翬ZZ"тTS,hmƦ*jea6e{~k*AŶh6o+lX$^\lql X%
hU64Akke,ݐ^0K3uɖhkjrVi@&