1/1+a pangkat n-m + 1/1+a pangkat m-n =.... Question from @ClaraNatacia

October 2018 1 185 Report

1/1+a pangkat n-m + 1/1+a pangkat m-n =...

izharulhaq78 $\displaystyle \frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{1}{1+\dfrac{a^{\displaystyle n}}{a^{\displaystyle m}}}+\frac{1}{1+\dfrac{a^{\displaystyle m}}{a^{\displaystyle n}}}\\\frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{1}{\dfrac{a^{\displaystyle m}+a^{\displaystyle n}}{a^{\displaystyle m}}}+\frac{1}{\dfrac{a^{\displaystyle n}+a^{\displaystyle m}}{a^{\displaystyle n}}}$
$\displaystyle \frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{1}{\dfrac{a^{\displaystyle m}+a^{\displaystyle n}}{a^{\displaystyle m}}}\times\frac{a^{\displaystyle m}}{a^{\displaystyle m}}+\frac{1}{\dfrac{a^{\displaystyle n}+a^{\displaystyle m}}{a^{\displaystyle n}}}\times\frac{a^{\displaystyle n}}{a^{\displaystyle n}}$
$\displaystyle \frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{a^{\displaystyle m}}{\left(\dfrac{a^{\displaystyle n}+a^{\displaystyle m}}{a^{\displaystyle m}}\right)a^{\displaystyle m}}+\frac{a^{\displaystyle n}}{\left(\dfrac{a^{\displaystyle n}+a^{\displaystyle m}}{a^{\displaystyle n}}\right)a^{\displaystyle n}}$
$\displaystyle \frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{a^{\displaystyle m}}{a^{\displaystyle m}+a^{\displaystyle n}}+\frac{a^{\displaystyle n}}{a^{\displaystyle n}+a^{\displaystyle m}}\\\frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=\frac{a^{\displaystyle m}+a^{\displaystyle n}}{a^{\displaystyle m}+a^{\displaystyle n}}$
$\displaystyle \boxed{\boxed{\frac{1}{1+a^{\displaystyle n-m}}+\frac{1}{1+a^{\displaystyle m-n}}=1}}$

ClaraNatacia kok bisa ditebalikkan ya? kurang ngerti di langkah ke 3

izharulhaq78 sudah diperjelas, coba dicek lagi