Ta có: \(S=1-3+3^2-3^3+....+3^{99}-3^{100}\)
\(3S=3-3^2+3^3-3^4+....+3^{100}-3^{101}\)
\(3S+S=\left(3-3^2+3^3-3^4+....+3^{100}-3^{101}\right)+\left(1-3+3^2-3^3+...+3^{99}-3^{100}\right)\)
\(4S=1-3^{100}\)
\(S=\frac{1-3^{100}}{4}\)
tìm x
a) -10-(x-5)+(3-x)=-8
b) 10+3(x-1)=10+6x
c) (x+1)(x-2)=0