例 題 


<例題>次の計算をせよ。

    (1)   (a1/2+a−1/2)    

     (2)   (a1/2−b1/2)(a1/2+b1/2)

    (3)   (a1/3+b1/3)(a2/3+a1/31/3+a2/3)

    (4)  (a1/4−b1/4)(a1/4+b1/4)(a1/2+b1/2)(a+b)

    (5)  (log102)+(log105)+(log105)(log108)

<解答>(1)   a1/2=x とおくと、a=x

         (a1/2+a−1/2)=(x+x−1)

                  =x+2xx−1+x−2

                  =x+2+x−2

                  =a+2+a−1



     (2)   a1/2=x, b1/2=y とおくと、a=x, b=y  

          (a1/2−b1/2)(a1/2+b1/2)=(x−y)(x+y)=x−y

                                                   =a−b



     (3)   a1/3=x,b1/3=y とおくと、a=x, b=y  

          (a1/3+b1/3)(a2/3−a1/31/3+a2/3)

                   =(x+y)(x−xy+y)

                   =x+y

                   =a+b



     (4)   a1/4=x,b1/4=y とおくと、

        a1/2=x, b1/2=y,a=y ,b=y  

          (a1/4−b1/4)(a1/4+b1/4)(a1/2+b1/2)(a+b)

                   =(x−y)(x+y)(x+y)(x+y)

                   =(x−y)(x+y)(x+y)

                   =(x−y)(x+y)

                   =x−y

                   =(a1/4)−(b1/4))

                   =a−b


    (5) log102=x、log105=y とおくと、

         x+y=log102+log105=log1010=1

            (log102)+(log105)+(log105)(log108)

              =(log102)+(log105)+3(log105)(log102)

              =x+y+3xy

              =(x+y)(x−xy+y)+3xy

              =(1)(x−xy+y)+3xy

              =x−xy+y+3xy

              =x+2xy+y

              =(x+y)

              =(1)

              =1


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