Lời giải:

\(\frac{1719}{3976}=\frac{1}{2+\frac{538}{1719}}=\frac{1}{2+\frac{1}{3+\frac{105}{538}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{13}{105}}}}=\frac{1}{2+\frac{1}{3+\frac{1}{5+\frac{1}{8+\frac{1}{13}}}}}\)

$\Rightarrow a=8; b=13$

\(\dfrac{1719}{3976}=\dfrac{1}{\dfrac{3976}{1719}}=\dfrac{1}{2+\dfrac{538}{1719}}=\dfrac{1}{2+\dfrac{1}{\dfrac{1719}{538}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{105}{538}}}\)

\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{\dfrac{538}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{13}{105}}}}=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{\dfrac{105}{13}}}}}\)

\(=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{5+\dfrac{1}{8+\dfrac{1}{13}}}}}\)

1719​=17193976​1​=2+1719538​1​=2+5381719​1​1​=2+3+538105​1​1​

=12+13+1538105=12+13+15+13105=12+13+15+110513=2+3+105538​1​1​1​=2+3+5+10513​1​1​1​=2+3+5+13105​1​1​1​1​

=12+13+15+18+113=2+3+5+8+131​1​1​1​1​
 

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1

mn mn ơiii

helllppppppppp

\(2,\\ 3^{n-3}+2^{n-3}+3^{n+1}+2^{n+2}\\ =3^{n-3}\left(1+3^4\right)+2^{n-3}\left(1+2^5\right)\\ =3^{n-3}\cdot82+2^{n-3}\cdot33\)

Vì \(3^{n-3}\cdot82⋮2;⋮3\) nên \(3^{n-3}\cdot82⋮6\)

\(2^{n-3}\cdot33⋮2;⋮3\) nên \(2^{n-3}\cdot33⋮6\)

Do đó tổng trên chia hết cho 6 với mọi \(n\in N\)

\(\dfrac{1}{a}-\dfrac{1}{b}=1\)

\(\Leftrightarrow\dfrac{b-a}{ab}=1\)

\(\Leftrightarrow b-a=ab\)

\(\Leftrightarrow a+ab-b=0\)

c)\(7^{2n}+7^{2n+2}=2450\)

⇒\(7^{2n}+7^{2n}.7^2=2450\)

⇒\(7^{2n}.50=2450\)

⇒\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

b: =>\(\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{n\left(n+1\right)}=\dfrac{200}{101}\)

=>\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{n\left(n+1\right)}=\dfrac{100}{101}\)

=>1-1/2+1/2-1/3+...+1/n-1/n+1=100/101

=>1-1/(n+1)=100/101

=>1/(n+1)=1/101

=>n+1=101

=>n=100