Ta có: \(\overline{aaa}=1+2+...+n=\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow\frac{n\left(n+1\right)}{2}=111a\Rightarrow n\left(n+1\right)=2.111a=2.3.37.a\)

Vì n(n+1) chia hết cho 37 nên một trong hai số chia hết cho 37

Mà \(\frac{n\left(n+1\right)}{2}\) là số có ba chữ số nên n và n+1 nhỏ hơn 74 => n=37 hoặc n+1=37

Nếu n=37 thì n+1=38 => \(\overline{aaa}=\frac{n\left(n+1\right)}{2}=\frac{37.38}{2}=703\) (loại)

Nếu n+1=37 thì n=36 => \(\overline{aaa}=\frac{n\left(n+1\right)}{2}=\frac{36.37}{2}=666\) (thỏa mãn)

Vậy n=36 và aaa = 666

a) 2/3=36/54  => x=36

b)15/6=10/3  =>x=3

\(a,\frac{2}{3}=\frac{x}{54}\Rightarrow x=\frac{2\times54}{3}\Rightarrow x=36\)

\(b,\frac{10}{x}=\frac{15}{6}\Rightarrow x=\frac{10\times6}{15}\Rightarrow x=4\)

\(c,\frac{2}{3}< \frac{x}{6}< 1\Rightarrow\frac{2}{3}\times6< x< 1\times6\Rightarrow4< x< 6\)

ta có:

1-1/2+1/2-1/3+1/3-1/4+....+1/x -1/x+1 =499/500

1-1/x+1 =499/500

1/x+1 =1/500 

x+1=500

x=499

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{X\times\left(X+1\right)}=\frac{499}{500}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{X}-\frac{1}{X+1}=\frac{499}{500}\)

\(\Leftrightarrow1-\frac{1}{X+1}=\frac{499}{500}\)

\(\Leftrightarrow\frac{1}{X+1}=\frac{1}{500}\)

\(\Leftrightarrow X+1=500\)

\(\Leftrightarrow X=499\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)

\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

a, 2.(x – 5)+7 = 77

<=> 2.(x – 5) = 70 <=> x – 5 = 35 <=> x = 40

b,  x - 1 3 - 3 5 : 3 4 + 2 . 2 3 = 14

<=> x - 1 3 - 3 + 2 4 = 14

<=>  x - 1 3 = 14 + 3 - 16 = 1

<=> x – 1 = 1 <=> x = 2

c,  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1

Đặt: A = 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 => 2A =  2 + 2 2 + 2 3 + . . . + 2 2017

=> 2A – A = ( 2 + 2 2 + 2 3 + . . . + 2 2017 ) – ( 1 + 2 + 2 2 + 2 3 + . . . + 2 2016 )

=> A =  2 2017 - 1

Ta có:  1 + 2 + 2 2 + 2 3 + . . . + 2 2016 = 2 x - 1 - 1 =>  2 2017 - 1 =  2 x - 1 - 1 => x = 2018

d,  5 2 x - 3 - 2 . 5 2 = 5 2 . 3

<=>  5 2 x - 3 = 5 2 . 3 + 5 2 . 2

<=>  5 2 x - 3 = 5 2 . ( 3 + 2 )

<=>  5 2 x - 3 = 5 3

<=> 2x – 3 = 3 => x = 3