\(\Leftrightarrow3B=3^2+3^3+...+3^{101}\\ \Leftrightarrow3B-B=3^{101}-3\\ \Leftrightarrow2B=3^{101}-3\\ \Leftrightarrow2B+3=3^{101}=3^n\\ \Leftrightarrow n=101\)

cảm ơn cậu nhìu lém

\(a,3⋮x-1\Rightarrow x-1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

.\(b,32⋮x\Rightarrow x\inƯ\left(32\right)=\left\{\pm1;\pm2;\pm4;\pm8;\pm16;\pm32\right\}\)

\(12⋮x\Rightarrow x\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

\(A=a^4-2a^3+3a^2-4a+5\)

   \(=\left(a^4-2a^3+a^2\right)+2\left(a^2-2a+1\right)+3\)

    \(=\left(a^2-a\right)^2+2\left(a-1\right)^2+3\ge3\)

Dấu "=" xảy ra <=>  a = 1

Vậy .......

a: \(A=3+3^2+\cdots+3^{100}\)

=>\(3A=3^2+3^3+\cdots+3^{101}\)

=>3A-A=\(3^2+3^3+\cdots+3^{101}-3-3^2-\cdots-3^{100}\)

=>\(2A=3^{101}-3\)

=>\(2A+3=3^{101}\)

=>\(3^{4n+1}=3^{101}\)

=>4n+1=101

=>4n=100

=>n=25

b: \(x^2+1=6y^2+2\)

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

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

=>\(y^2=\frac{x^2-1}{6}\)

=>\(y^2\) chẵn

=>y chẵn

mà y là số nguyên tố

nên y=2

\(x^2-6y^2=1\)

=>\(x^2=6y^2+1=6\cdot2^2+1=6\cdot4+1=24+1=25\)

=>x=5(nhận)

a: \(A=3+3^2+\cdots+3^{100}\)

=>\(3A=3^2+3^3+\cdots+3^{101}\)

=>\(3A-A=3^2+3^3+\cdots+3^{101}-3-3^2-\cdots-3^{100}\)

=>\(2A=3^{101}-3\)

=>\(2A+3=3^{101}\)

=>\(3^{4n+1}=3^{101}\)

=>4n+1=101

=>4n=100

=>n=25

b: \(x^2+1=6y^2+2\)

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

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

=>\(y^2=\frac{x^2-1}{6}\)

=>\(y^2\) ⋮2

=>y⋮2

mà y là số nguyên tố

nên y=2

\(x^2-6y^2=1\)

=>\(x^2=6y^2+1=6\cdot2^2+1=6\cdot4+1=24+1=25=5^2\)

=>x=5

\(a,A=3+3^2+3^3+3^4+...+3^{100}\\ 3A=3^2+3^3+3^4+3^5+3^{101}\\ 3A-A=2A=3^{101}-3\\ \Rightarrow2A+3=3^{101}=3^{4.25+1}\\ \Rightarrow n=25\)

Giải:

Vì 2a36b chia hết cho 2;3;5 nên b = 0

Do a360 \(⋮\)3, ta có:

2+a+3+6+0 = a + 11

 =>a+11 \(⋮\)3( a<10)

 => a = 1;4;7

Tổng của các chữ số a là: 1+4+7=12

Đáp số: 12

A = 235

Tck mk nhé !

sao lại 2 lần 31 the kia !!!!