vì số đó chia cho 11 dư 6, chia cho 17 dư 12 chia cho 29 dư 24 nên khi thêm 5 đơn vị vào số đó ta được số chia hết cho cả 11; 17; 29

BCNN(11; 17;29) = 11 x 17 x 29 = 5423

số cần tìm là 5423 - 5 = 5418

Gọi số cần tìm là \(x\left(x\inℕ\right)\)

Theo đề cho, ta có: \(x-15⋮20,25,30\Rightarrow x-15\in BC\left(20,25,30\right)\)

\(\rightarrow\)Ta có:

\(20=2^2.5\)

\(25=5^2\)

\(30=2.3.5\)

\(\Rightarrow BCNN\left(20,25,30\right)=2^2.5^2.3=300\)

\(\Rightarrow BC\left(20,25,30\right)=B\left(300\right)=\left\{0;300;600;...\right\}\)

\(\Rightarrow x-15\in\left\{0;300;600;...\right\}\)

\(\Rightarrow x\in\left\{15;315;615;...\right\}\)

Mà đề cho: x là số tự nhiên có ba chữ số chia hết cho 41 \(\Rightarrow x=615\)

`6.x-5=613`

`6.x=613+5=618`

`x=618:6=103`

=> 6x = 613 + 5 = 618

=> x = 618 : 6 = 103

Câu 1.

a) \(-3\left|x-6\right|+53=27\)

\(\Rightarrow-3\left|x-6\right|=-26\\ \Rightarrow\left|x-6\right|=\dfrac{26}{3}\\ \Rightarrow\left[{}\begin{matrix}x-6=\dfrac{26}{3}\\x-6=-\dfrac{26}{3}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{44}{3}\\x=-\dfrac{8}{3}\end{matrix}\right.\left(L\right)\)

Vậy không có \(x\in Z\) thỏa mãn

b) \(\left(3-x\right)\left(x-8\right)>0\)

TH1: \(\left\{{}\begin{matrix}3-x>0\\x-8>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x>8\end{matrix}\right.\left(L\right)\)

TH2: \(\left\{{}\begin{matrix}3-x< 0\\x-8< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x< 8\end{matrix}\right.\Rightarrow3< x< 8\)

Mà \(x\in Z\Rightarrow x\in\left\{4;5;6;7\right\}\)

`1/20+1/30+1/42+1/56+1/72+1/90+1/110+1/132`

`=1/[4.5]+1/[5.6]+1/[6.7]+1/[7.8]+1/[8.9]+1/[9.10]+1/[10.11]+1/[11.12]`

`=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12`

`=1/4-1/12`

`=3/12-1/12=2/12=1/6`

\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) +.......\(\dfrac{2}{x\left(x+2\right)}\) = \(\dfrac{100}{101}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +.......\(\dfrac{1}{x}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{100}{101}\)

\(\dfrac{1}{1}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{100}{101}\) 

\(\dfrac{1}{x+2}\) = \(\dfrac{1}{1}\) - \(\dfrac{100}{101}\)

\(\dfrac{1}{x+2}\) = \(\dfrac{1}{101}\)

x + 2 = 101

 x= 101 - 2

x = 99

vậy x ϵ {99}

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{100}{101}\)

Điều kiện: \(x\ne0;x\ne-2\)

\(\Rightarrow1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{100}{101}\)

\(\Rightarrow1-\dfrac{1}{x+2}=\dfrac{100}{101}\)

\(\Rightarrow\dfrac{1}{x+2}=\dfrac{1}{101}\)

\(\Rightarrow x+2=101\)

\(\Rightarrow x=99\)

A = 2011.2013 =  2011 . (2012+1)  =2012 . 2011 + 2011

B = 2012 .2012 = 2012 . (2011 + 1) = 2012 . 2011 + 2012

A = 2012. 2011 + 2011 < 2012 . 2011 + 2012 = B

vậy A < B 

Biểu thức B có vẻ sai. Bạn xem lại.

A=  { 0; 1; 2; 4; 5}

A = {xϵN| x<6 }

B = {1; 2; 3; 4; 5; .......29}

B = {x ϵ N*| 0<x<30}

C = {0; 18; 36; 54}

C = {x = 18k| kϵ N; 0≤ k≤3}

  hi

A = {5; 6; 7}

A = { x ϵ N| 4<x ≤7}

B = {1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12}

B = { x ϵ N*| 0< x ≤12}