Bài 1:

$1+2+3+...+n=\overline{aaa}$

$\Leftrightarrow \frac{n(n+1)}{2}=a.111$

$\Leftrightarrow n(n+1)=a.222\vdots 37$ nên suy ra $n\vdots 37$ hoặc $n+1\vdots 37$

Nếu $n\vdots 37$. Đặt $n=37k$ với $k\in\mathbb{N}^*$

Khi đó: $37k(37k+1)=222a\Rightarrow k(37k+1)=6a$

$6a\leq 54$ do $a\leq 9; 37k+1\geq 38$ do $k\geq 1$

$\Rightarrow k=\frac{6a}{37k+1}< 2\Rightarrow k=1$

$\Rightarrow 6a=38$ (vô lý)

Nếu $n+1\vdots 37$. Đặt $n+1=37k$ với $k\in\mathbb{N}^*$

Khi đó: $(37k-1).37k=222a\Rightarrow k(37k-1)=6a$

$6a\leq 54$ do $a\leq 9$; $37k-1\geq 36$ do $k\geq 1$

$\Rightarrow k=\frac{6a}{37k-1}< 2\Rightarrow k=1$

$\Rightarrow n=36; a=6$

Bài 2: $5z=7z$ hình như sai, bạn coi lại đề.

Bài 3:

\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Leftrightarrow \frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)

\(\Leftrightarrow \frac{9a+(a+b)}{a+b}=\frac{9b+(b+c)}{b+c}\Leftrightarrow \frac{9a}{a+b}+1=\frac{9b}{b+c}+1\)

\(\Leftrightarrow \frac{a}{a+b}=\frac{b}{b+c}\Rightarrow ab+ac=ab+b^2\)

\(\Leftrightarrow ac=b^2\Rightarrow \frac{a}{b}=\frac{b}{c}\) (đpcm)

Bài 2 sau khi đã sửa đề thành $5x=7z$:

Ta có:
\(\frac{x}{y}=\frac{3}{2}\Leftrightarrow \frac{x}{3}=\frac{y}{2}\Leftrightarrow \frac{x}{21}=\frac{y}{14}(1)\)

\(5x=7z\Leftrightarrow \frac{x}{7}=\frac{z}{5}\Leftrightarrow \frac{x}{21}=\frac{z}{15}(2)\)

Từ $(1);(2)\Rightarrow \frac{x}{21}=\frac{y}{14}=\frac{z}{15}$ và đặt bằng $k$

$\Rightarrow x=21k; y=14k; z=15k$

Khi đó:

$x-2y+z=32$

$\Leftrightarrow 21k-28k+15k=32\Leftrightarrow 8k=32\Rightarrow k=4$

$\Rightarrow x=21k=84; y=14k=56; z=15k=60$

a: =>y+1,2=(5,34-10,34):5=-1

=>y=-2,2

b: =>x(x+1)/2=111a

=>x(x+1)=222a

=>\(x\in\varnothing\)

1) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Rightarrow3^x\left(1+3^1+3^2\right)=351\)

\(\Rightarrow3^x.13=351\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

2) \(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

\(\Rightarrow C=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)...+2^{96}\left(2+2^2+2^3+2^4\right)\)

\(\Rightarrow C=30+2^4.30...+2^{96}.30\)

\(\Rightarrow C=\left(1+2^4+...+2^{96}\right).30⋮30\)

mà \(30=5.6\)

\(\Rightarrow C⋮5\left(dpcm\right)\)

1,

Có \(3^x\)+ \(3^{x+1}\) + \(3^{x+2}\) = \(351\)

=> \(3^x\) + \(3^x\).\(3\) + \(3^x\).\(9\) = \(351\)

=> \(3^x\).\(13\) = \(351\)

=> \(3^x\) = \(27\)

=> \(x\) = \(3\)

2,

C = \(2\) + \(2^2\) + \(2^3\) + ... + \(2^{100}\)

2C = \(2^2\) + \(2^3\) + \(2^4\) + ... + \(2^{101}\)

2C - C = \(2^{101}\) - \(2\)

C = \(2^{101}\) - \(2\)

C = \(2\).\(\left(2^{100}-1\right)\)

C = 2.\(\left(\left(2^5\right)^{20}-1^{20}\right)\)

Có \(2^5\) \(-1\) \(⋮\) 5

=> \(\left(\left(2^5\right)^{20}-1^{20}\right)\) \(⋮\) 5

=> C \(⋮\) 5

3,

Xét \(\overline{abcdeg}\)

= \(\overline{ab}\).\(10000\) + \(\overline{cd}\).\(100\) + \(\overline{eg}\)

= \(\left(\overline{ab}+\overline{cd}+\overline{eg}\right)\) + \(9.\left(1111.\overline{ab}+11.\overline{cd}\right)\)

Có\(\left\{{}\begin{matrix}9.\left(1111.\overline{ab}+11.\overline{cd}\right)⋮9\left(1111.\overline{ab}+11.\overline{cd}\inℕ^∗\right)\\\overline{ab}+\overline{cd}+\overline{eg}⋮9\end{matrix}\right.\)

=> \(\overline{abcdeg}⋮9\)

4,

S = \(3^0+3^2+3^4+...+3^{2002}\)

9S = \(3^2+3^4+3^6+...+3^{2004}\)

9S - S = \(3^2+3^4+3^6+...+3^{2004}\) - (\(3^0+3^2+3^4+...+3^{2002}\))

8S = \(3^{2004}-1\)

=> 8S \(< 3^{2004}\)

 1 + 2 + 3 + ... + n = \(\overline{aaa}\)

=> ( n + 1 ) x n : 2 = 3 x 37 x a

=> n x ( n + 1 ) = 6a x 37

Vì n x ( n + 1 ) là tích 2 số liên tiếp nên 6a x 37 là tích 2 số tự nhiên liên tiếp 

=> 6a = 36

=> a = 6 ( vì a \(\in\) N )

Do đó n x ( n + 1 ) = 36 x 37

=> n = 36 ( vì n \(\in N\)*)

Vậy n = 36; a = 6

a)Ta thấy: \(\left\{{}\begin{matrix}\left(12x-y+7\right)^{2016}\ge0\forall x,y\\\left|2x-3\right|^{2017}\ge0\forall x\end{matrix}\right.\)

\(\Rightarrow\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}\ge0\forall x,y\)

Mà \(\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}\le0\)

Nên xảy ra khi \(\left(12x-y+7\right)^{2016}+\left|2x-3\right|^{2017}=0\)

\(\left\{{}\begin{matrix}\left(12x-y+7\right)^{2016}=0\\\left|2x-3\right|^{2017}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}12x-y+7=0\\x=\dfrac{3}{2}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}y=25\\x=\dfrac{3}{2}\end{matrix}\right.\)

b)\(1+2+3+...+n=\overline{aaa}\)

Ta có: \(\left\{{}\begin{matrix}VT=\dfrac{n\left(n+1\right)}{2}\\VP=a\cdot111\end{matrix}\right.\)

\(\Rightarrow\dfrac{n\left(n+1\right)}{2}=a\cdot111\Rightarrow n\left(n+1\right)=a\cdot222\)

\(\Rightarrow n\left(n+1\right)=6a\cdot37=6a\left(36+1\right)\)

Dễ thấy: \(n\left(n+1\right)\) là \(2\) số tự nhiên liên tiếp và \(6a\) và \(36+1\) là 2 số tự nhiên liên tiếp

\(\Rightarrow6a=36\Rightarrow a=6\Rightarrow n=36\)

X. ( X - 1) . X  ( X - 1 ) = ( X - 2) XX ( X -  1) 

X . X - X . 1 . X . X - X . 1 = X . X - X . 2 . X . X - X . 1 

2X - X . 1 . 2X - X . 1 = 2X - X. 2 . 2X - X 

2 . 1 . 2 . 1 = 2 . 2 . 1 

4               = 4 

b: Ta có: \(3^x+2\cdot3^{x-2}=297\)

\(\Leftrightarrow3^x=297:\dfrac{11}{9}=243\)

hay x=5