Bài 1 :

Gọi \(A=5+5^2+5^3+...+5^{98}+5^{99}\\ 5A=5^2+5^3+5^4+...+5^{99}+5^{100}\\ 5A-A=\left(5^2+5^3+5^4+...+5^{99}+5^{100}\right)-\left(5+5^2+5^3+...+5^{98}+5^{99}\right)\\ 4A=5^{100}-5\\ A=\dfrac{5^{100}-5}{4}\)

Bài 2:

\(\left(12x-4\right)\cdot8^{2022}=4\cdot8^{2023}\\ 12x-4=4\cdot8^{2023}:8^{2022}\\ 12x-4=4\cdot8\\ 12x-4=32\\ 12x=36\\ x=3\)

;-; oce ạ

Có \(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\\ =\dfrac{y+x}{y}\cdot\dfrac{z+y}{z}\cdot\dfrac{x+z}{x}\)

Áp dụng t/c của DTSBN , ta đc :

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ = \dfrac{y+z-x+z+x-y+x+y-z}{x+y+z}\\ =\dfrac{\left(y+y-y\right)+\left(x+x-x\right)+\left(z+z-z\right)}{x+y+z}\\ =\dfrac{x+y+z}{x+y+z}=1\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{y+z-x}{x}=1\\\dfrac{z+x-y}{y}=1\\\dfrac{x+y-z}{z}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z-x=x\\z+x-y=y\\x+y-z=z\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}y+z=2x\\z+x=2y\\x+y=2z\end{matrix}\right.\left(1\right)\)

Thay `(1)` vào `P` ta có :

\(P=\dfrac{2z}{y}\cdot\dfrac{2x}{z}\cdot\dfrac{2y}{x}=\dfrac{8xyz}{xyz}=8\)

ủa :DDD , khó v cx nghĩ ra