a) Khi vẽ n tia chung góc thì số góc tạo thành là \(\dfrac{n\left(n-1\right)}{2}\) góc 

b) Khi số góc tạo thành là 10 góc thì ta có:

\(\dfrac{n\left(n-1\right)}{2}=10\)

\(\Rightarrow n\left(n-1\right)=2\cdot10=20\)

\(\Rightarrow n\left(n-1\right)=5\cdot4=5\cdot\left(5-1\right)\)

\(\Rightarrow n=5\) 

màu hồng

màu hồng

Ngu

36 x 28 + 64 x 59 - 36 x 128 + 32 x 82

36 x 28 + 32 x 2 x 59 - 36 x 128 + 32 x 82

36 x 28 + 32 x 118 - 36 x 128 + 32 x 82

36 x (28 + 128) + 32 x (82 +118)

36 x 156 + 32 x 200

Tự tính nốt

\(\dfrac{3}{4}\cdot2024^0-\left(\dfrac{13}{11}-\dfrac{1}{2}\right):\dfrac{2}{11}\)

\(=\dfrac{3}{4}\cdot1-\dfrac{15}{22}:\dfrac{2}{11}\)

\(=\dfrac{3}{4}-\dfrac{15}{22}\cdot\dfrac{11}{2}\)

\(=\dfrac{3}{4}-\dfrac{15}{4}\) 

\(=\dfrac{-12}{4}=-3\)

\(-\dfrac{2}{3}+\dfrac{6}{5}:\dfrac{2}{3}+\left(\dfrac{-2}{15}\right)=\dfrac{2}{3}+\left(-\dfrac{6}{5}\right):\dfrac{2}{3}+\left(\dfrac{-2}{15}\right)=\dfrac{2}{3}:\left(-\dfrac{6}{5}+\dfrac{-2}{15}\right)=\dfrac{2}{3}:\left(-\dfrac{18}{15}+\dfrac{-2}{15}\right)=\dfrac{2}{3}:\dfrac{-20}{15}=\dfrac{2}{3}X\dfrac{15}{-20}=\dfrac{1}{1}x\dfrac{5}{-10}=\dfrac{5}{-10}=\dfrac{1}{-2}\)

\(\dfrac{-2}{3}+\dfrac{6}{5}:\dfrac{2}{3}+\dfrac{-2}{15}\)

\(=\dfrac{-2}{3}+\dfrac{6}{5}\times\dfrac{3}{2}-\dfrac{2}{15}\)

\(=\dfrac{-2}{3}+\dfrac{9}{5}-\dfrac{2}{15}\)

\(=\dfrac{-10}{15}+\dfrac{27}{15}-\dfrac{12}{15}\)

\(=\dfrac{5}{15}=\dfrac{1}{3}\)

11,2 > 8,1 

- 11,2 < - 8,1

11,2 > 8,1

11,2 x (-1) < 8,1 x (-1)

- 11,2 < - 8,1

Vậy  -11,2 < - 8,1 

\(-\dfrac{20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}=\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\left(\dfrac{2}{5}+\dfrac{7}{15}\right)+\dfrac{2}{3}=-1+\dfrac{13}{15}+\dfrac{2}{3}=\dfrac{-2}{15}+\dfrac{2}{3}=\dfrac{-2}{15}+\dfrac{10}{15}=\dfrac{8}{15}\)

-20/23+2/3-3/23+2/5+7/15

=(-20/13-3/23)+(2/3+2/5+7/15)

=(-1)+(10/15+6/15+7/15)

=(-1)+23/15

=8/16

\(\dfrac{2}{3}+\dfrac{-3}{4}+\dfrac{5}{6}\)

\(=\dfrac{8}{12}+\dfrac{-9}{12}+\dfrac{10}{12}\)

\(=\dfrac{8}{12}-\dfrac{9}{12}+\dfrac{10}{12}\)

\(=\dfrac{-1}{12}+\dfrac{10}{12}\)

\(=\dfrac{9}{12}=\dfrac{3}{4}\)

\(\dfrac{2}{3}+-\dfrac{3}{4}+\dfrac{5}{6}=\dfrac{8}{12}+\left(-\dfrac{9}{12}\right)+\dfrac{10}{12}=\dfrac{8+\left(-9\right)+10}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

Gọi 8 số đó là \(n_i=\overline{a_ib_i}\) với \(1\le i\le8\).

Với mỗi 2 số \(n_i,n_j\left(i\ne j,1\le i,j\le8\right)\), ta có:

\(N_{ij}=\overline{a_ib_i0a_jb_j}\) 

\(=10000a_i+1000b_i+10a_j+b_j\)

\(=10010a_i+1001b_i+\left(10a_j-10a_i\right)+\left(b_j-b_i\right)\)

\(=10010a_i+1001b_i+n_j-n_i\)

 Để ý rằng một số khi chia cho 7 chỉ có 7 số dư phân biệt là 0, 1, 2,..., 6. Do ta chọn 8 số \(n_i\) nên theo nguyên lý Dirichlet sẽ tồn tại 2 số \(n_k,n_l\left(k\ne l,1\le k,l\le8\right)\) mà chúng có cùng số dư khi chia cho 7.

 \(\Rightarrow n_k-n_l⋮7\)

 Khi đó \(N_{kl}=10010a_k+1001b_k+\left(n_l-n_k\right)⋮7\)  (do \(1001⋮7\))

 Vậy ta có đpcm.

đầu bài là j vậy ạ

\(3\dfrac{1}{5}+\dfrac{2}{5}:\dfrac{-2}{3}\)
\(=\dfrac{16}{5}+\dfrac{2}{5}.\dfrac{-3}{2}\)
\(=\dfrac{16}{5}+\dfrac{-3}{5}\)
\(=\dfrac{13}{5}\)