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

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​
 

a: ĐKXĐ: \(x\in R\)

\(B=\dfrac{x^2+15}{x^2+3}\)

\(=\dfrac{x^2+3+12}{x^2+3}\)

\(=1+\dfrac{12}{x^2+3}\)

\(x^2+3>=3\forall x\)

=>\(\dfrac{12}{x^2+3}< =\dfrac{12}{3}=4\forall x\)

=>\(\dfrac{12}{x^2+3}+1< =5\forall x\)

=>\(B< =5\forall x\)

Dấu '=' xảy ra khi x=0

aaaassssssssssssssssssssddddddddddd

\(a,A=\dfrac{-3\left(2n-3\right)-8}{2n-3}=-3-\dfrac{8}{2n-3}\in Z\\ \Leftrightarrow2n-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow n\in\left\{1;2\right\}\left(n\in Z\right)\)

\(b,\dfrac{ab}{a+2b}=\dfrac{3}{2}\Leftrightarrow\dfrac{a+2b}{ab}=\dfrac{2}{3}\Leftrightarrow\dfrac{1}{b}+\dfrac{2}{a}=\dfrac{2}{3}\\ \dfrac{bc}{b+2c}=\dfrac{4}{3}\Leftrightarrow\dfrac{b+2c}{bc}=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{c}+\dfrac{2}{b}=\dfrac{3}{4}\\ \dfrac{ca}{c+2a}=3\Leftrightarrow\dfrac{c+2a}{ca}=\dfrac{1}{3}\Leftrightarrow\dfrac{1}{a}+\dfrac{2}{c}=\dfrac{1}{3}\)

Cộng vế theo vế \(\Leftrightarrow\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}=\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{7}{4}\)

\(\Leftrightarrow3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{7}{4}\\ \Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{7}{12}\\ \Leftrightarrow\dfrac{ab+bc+ca}{abc}=\dfrac{7}{12}\\ \Leftrightarrow T=\dfrac{12}{7}\)

a) Ta có: \(\dfrac{3+x}{7+y}=\dfrac{3}{7}\)

\(\Leftrightarrow\dfrac{x+3}{3}=\dfrac{y+7}{7}\)

mà x+y=20

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x+3}{3}=\dfrac{y+7}{7}=\dfrac{x+y+3+7}{3+7}=\dfrac{20+10}{10}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x+3}{10}=3\\\dfrac{y+7}{7}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3=30\\y+7=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=14\end{matrix}\right.\)

Vậy: x=27; y=14

Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)

\(\dfrac{138}{30}< X< \dfrac{200}{3}\)

\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)

Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)

\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)

\(\Rightarrow X=\dfrac{105381}{52}\)