Xin đề bài 1b

1.D

2.D

3.B

1.D

2.D

3.B

2a+8+1b/3

=2.2+8+1.3/3

=4+8+1

=12+1

=13

\(\frac{\text{2a+8 +1b }}{3}\) . Thay a=2; b=3 .Ta có

\(\frac{2.2+8+1.3}{3}\)=\(\frac{15}{3}\)=5

a) Để (d) song song với đường thẳng y = x - 1.

=> 3 - m = 1.

<=> m = 2.

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

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