4^x-1+1=65​

4^x-1    =65-1

4^x-1     =64

4^x-1     =4^3

x-1         =3

x             =3-1

x              =2

Vậy,x= 2

4^x-1+1=65​

4^x-1    =65-1

4^x-1     =64

4^x-1     =4^3

x-1         =3

x             =3-1

x              =2

Vậy,x= 2

Cũng được

Lời giải:

$A=9^0+(9+9^2)+(9^3+9^4)+....+(9^{2013}+9^{2014})$

$=1+9(1+9)+9^3(1+9)+....+9^{2013}(1+9)$
$=1+(1+9)(9+9^3+....+9^{2013})$

$=1+10(9+9^3+....+9^{2013})$

$\Rightarrow A$ chia $10$ dư $1$.

** Bổ sung điều kiện $x,y$ là số nguyên.

$x^2+(y+1)^2=1$

$\Rightarrow x^2=1-(y+1)^2\leq 1$ (do $(y+1)^2\geq 0$)

$\Rightarrow -1\leq x\leq 1$

Mà $x$ nguyên nên $x\in \left\{-1; 0; 1\right\}$.

Nếu $x=0$ thì $(y+1)^2=1-x^2=1\Rightarrow y+1=\pm 1\Rightarrow y=0$ hoặc $y=-2$

Nếu $x=-1$ thì $(y+1)^2=1-x^2=0\Rightarrow y=-1$

Nếu $x=1$ thì $(y+1)^2=1-x^2=0\Rightarrow y=-1$

S=1+3+3^2+...+3^100

3S=3+3^2+3^3+...+3^101

3S-S=(3+3^2+3^3+...+3^101)-(1+3+3^2+...+3^100)

2S=3+3^2+3^3+...+3^101-1-3-3^2-...-3^100

2S=3^101-1

Suy ra:2S+1=3^101-1+1=3^101

Mà 2S+1=3^n=3^101 nên n=101

n= 101

nho k minh nha

1999 . 47 + 1999 . 53 

                 = 1999 . ( 47 + 53 ) 

                = 1999 . 100 

                 = 199900

=1999.(47+53)

=1999.100

=199900

a/

$x+4\vdots x+1$
$\Rightarrow (x+1)+3\vdots x+1$
$\Rightarrow 3\vdots x+1$

$\Rightarrow x+1\in \left\{1; 3\right\}$

$\Rightarrow x\in \left\{0; 2\right\}$

b/

$10\equiv 1\pmod 9$

$\Rightarrow 10^n\equiv 1^n\equiv 1\pmod 9$

$5^3=125\equiv 8\pmod 9$

$\Rightarrow 10^n+5^3\equiv 1+8\equiv 0\pmod 9$

$\Rightarrow 10^n+5^3\vdots 9$

Vì $10^n+5^3\vdots 9; 9\vdots 3\Rightarrow 10^n+5^3\vdots 3$.

=(3+3²+3³)+(3⁴+3⁵⁺3⁶)+.....+(3⁷+3⁸+3⁸)

=3(1+3+3²)+3⁴(1+3+3²)+.....+3⁷(1+3+3²)

=3.13+3⁴.13+.....+3⁷.13

=13(3+3⁴+....+3⁷) se chia het cho 13

\(A=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)

\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)

\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)

\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)