Lời giải:
Nếu $n$ chia hết cho $3$. Đặt $n=3k$ với $k$ tự nhiên.

Khi đó: $A=10^n+18n-1=10^{3k}+18.3k-1=1000^k+54k-1$

Có:
$1000\equiv 1\pmod {27}\Rightarrow 1000^k\equiv 1^k\equiv 1\pmod {27}$

$54k\equiv 0\pmod {27}$

$\Rightarrow 1000^k+54k-1\equiv 1+0-1\equiv 0\pmod {27}$

Hay $A\equiv 0\pmod {27}(1)$

Nếu $n$ chia $3$ dư $1$. Đặt $n=3k+1$ với $k$ tự nhiên.

Khi đó:

$A=10^{3k+1}+18(3k+1)-1=1000^k.10+54k+17$

$\equiv 1^k.10+0+17=27\equiv 0\pmod {27}(2)$

Nếu $n$ chia $3$ dư $2$. Đặt $n=3k+2$ với $k$ tự nhiên.

Khi đó:

$A=10^{3k+2}+18(3k+2)-1=1000^k.100+54k+35$

$\equiv 1^k.100+0+35=135\equiv 0\pmod {27}(3)$
Từ $(1); (2); (3)\Rightarrow A\vdots 27$ với mọi $n$ tự nhiên.

Em cảm ơn thầy/cô nhiều ạ .

B1:

\(B\left(7\right)=\left\{0;\pm7;\pm14;....\right\}\\ B\left(-7\right)=\left\{0;\pm7;\pm14;....\right\}\\ B\left(5\right)=\left\{0;\pm5;\pm10;...\right\}\\ B\left(-8\right)=\left\{0;\pm8;\pm16;...\right\}\)

B2:

\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\\ Ư\left(-12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\\ Ư\left(36\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm9;\pm12;\pm18;\pm36\right\};Ư\left(-8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

giúp mik vs 2x+5=x-1  mik bí lắm r

Bài 1:

ta có:\(\hept{\begin{cases}a.b=18\\a+b=11\end{cases}}\)

pt (1) \(\Leftrightarrow b\left(11-b\right)=18\)

\(\Rightarrow11b-b^2=18\)

\(\Rightarrow-\left(11b+b^2\right)=18\)

\(\Rightarrow b^2+11b=18\)

\(\Rightarrow b^2+11b+18=0\)

\(\Rightarrow b^2+11b+22-4=0\)

\(\Rightarrow\left(b+11\right)-4=0\)

\(\Rightarrow\left(b+11+2\right).\left(b+11-2\right)=0\)

\(\Rightarrow\left(b+13\right).\left(b+9\right)=0\)

\(\Rightarrow\hept{\begin{cases}b=-13\Leftrightarrow a=2\\b=-9\Leftrightarrow a=-2\end{cases}}\)

Bài 2:

2)

P=-[(2x-6)2+/-5-y/-37]

Vi (2x-6)2 ≥0 , /-5-y/≥0 nên (2x-6)2+/-5-y/-37-37/ => P≤37

Dấu = xảu ra khi x=3 , y=-5

Vậy Max P=37 khi x=3 , y=-5

chúc bạn học tốt !

=205 mà đề là tính hả

80 - (-5)³

= 80+ 5³

= 80+ 125

=205

B1

Số nhóm biểu thức nhò là

(2001+3)/2+1=1003(số)
mà giá trị mỗi biểu thức là 1

=> 1+1*1003=1004

Bài 1:

a) \(\dfrac{4}{5}-\dfrac{7}{6}+\dfrac{-6}{15}=\dfrac{24}{30}-\dfrac{35}{30}+\dfrac{-12}{30}=\dfrac{24-35+-12}{30}=\dfrac{-23}{30}\) 

b) \(\dfrac{-5}{9}.\dfrac{7}{13}+\dfrac{6}{13}.\dfrac{-5}{9}+3\dfrac{7}{9}\) 

\(=\dfrac{-5}{9}.\left(\dfrac{7}{13}+\dfrac{6}{13}\right)+\dfrac{34}{9}\) 

\(=\dfrac{-5}{9}.1+\dfrac{34}{9}\) 

\(=\dfrac{-5}{9}+\dfrac{34}{9}\) 

\(=\dfrac{29}{9}\) 

c) \(6\dfrac{3}{8}-\left(4\dfrac{3}{8}-\dfrac{1}{2}\right)=\dfrac{51}{8}-\dfrac{35}{8}+\dfrac{1}{2}=\left(\dfrac{51}{8}-\dfrac{35}{8}\right)+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\) 

d) \(2\dfrac{1}{3}.1,5-\left(\dfrac{11}{10}+50\%\right):\dfrac{4}{15}\) 

\(=\dfrac{7}{3}.1,5-\dfrac{8}{5}:\dfrac{4}{15}\) 

\(=\dfrac{7}{2}-6\) 

\(=\dfrac{-5}{2}=-2,5\) 

đề đâu mà làm:))