Hai kalian yang baik tolong bantu 2 nomor.. makasih

Hai kalian yang baik tolong bantu 2 nomor.. makasih

Jawaban Terkonfirmasi

Nomer 1

$$begin{align} left[frac{1}{1+p}right]^5.left[frac{1}{1-p}right]^{-7}.left[frac{p-1}{1+p}right]^{-6}~&=~left[frac{1}{1+p}right]^5~.~left[frac{1-p}{1}right]^{7}~.~left[frac{1+p}{p-1}right]^{6} \ &=~frac{1}{(1+p)^5}~times~(1-p)^7~times~frac{(1+p)^6}{(p-1)^6} \ &=~frac{1}{(1+p)^5}~times~(-1)(p-1)^7~times~frac{(1+p)^6}{(p-1)^6}end{align}$
$$begin{align} &=~frac{1}{not(1+p)^5}~times~frac{^1not(1+p)^6}{not(p-1)^6}~times~(-1)~~~^1not(p-1)^7 \ &=~(1+p)~times~(-1)~times~(p-1) \ &=~(1+p)~times~(1-p) \ &=~1-p^2end{align}$

Nomer 2

$$begin{align} a^2+b^2~~&=~~(a+b)^2~-~2ab \ 2~~&=~~1~-~2ab \ 2-1~~&=~~-2ab \ 1~~&=~~-2ab \ -frac{1}{2}~~&=~~ab \ \&boxed{a^4~+~b^4~=~(a+b)^4~-4a^3b-6a^2b^2-4ab^3}\$maka~:$ \ a^4+b^4~&=~1~-~4a^2(-frac{1}{2})-6(-frac{1}{2})^2-4b^2(-frac{1}{2}) \ &=~1~+2a^2-frac{3}{2}+2b^2 \ &=~-frac{1}{2}~+~2a^2+2b^2 \ &=~~-frac{1}{2}+2(a^2+b^2) \ &=~-frac{1}{2}+2(2) \ &=~-frac{1}{2}+4 \ &=~3frac{1}{2}end{align}$