(√3+√2)^[2log(√3-√2)(√5)]=(√3-√2)^[log(√3-√2)(1/5)]=1/5√3+√2=(√3-√2)^(-1)2log(√3-√2)(√5)=log(√3-√2)(5)则原式=(√3-√2)^[-log(√3-√2)(5)]=(√3-√2)^[log(√3-√2)(1/5)=1/5这里用到了公式a^[loga(x)]=x
化简，还是计算结果
