Answer:

a) 

Có:

\(15=1.15=\left(-1\right).\left(-15\right)=3.5=\left(-3\right).\left(-5\right)\)

b) 

Có:

\(M=1+3+3^2+3^3+3^4+...+3^{98}+3^{99}+3^{100}\)

\(=\left(1+3\right)+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2.\left(1+3+3^2\right)+...+3^{98}.\left(1+3+3^2\right)\)

\(=4+3^2.13+3^{98}.13\)

\(=4+13.\left(3^2+...+3^{98}\right)\)

Vậy M chia 13 dư 4

Có:

\(M=1+3+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3.40+3^5.40+...+3^{97}.40\)

\(=1+40.\left(3+3^5+...+3^{97}\right)⋮40\)

Mà \(40\left(3+3^5+...+3^{97}\right)⋮40\)

Vậy M chia 40 dư 1

Cíu ét o ét

Trả lời:

Ta có: \(\left(x-1\right)^2+5y^2=6\)

=>\(5y^2\le6\) và \(5y^2\) ⋮5

=>\(5y^2\in\left\lbrace0;5\right\rbrace\)

=>\(y^2\in\left\lbrace0;1\right\rbrace\)

TH1: \(y^2=0\)

Ta có: \(\left(x-1\right)^2+5y^2=6\)

=>\(\left(x-1\right)^2=6\)

mà x là số nguyên

nên x∈∅

TH2: \(y^2=1\)

Ta có: \(\left(x-1\right)^2+5y^2=6\)

=>\(\left(x-1\right)^2=6-5y^2=1\)

=>\(\left[\begin{array}{l}x-1=1\\ x-1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\left(nhận\right)\\ x=0\left(nhận\right)\end{array}\right.\)

\(y^2=1\)

=>y=1(nhận) hoặc y=-1(nhận)