Tổng A và B không chia hết cho 6 con tổng C thì chia hết cho 6

tong a va b

Gỉa sử không có a điểm nào thẳng hàng thì vẽ được : 20.19 : 2 = 190 (đường thẳng)

Số đường thẳng dôi ra là : 190 - 170 = 20 (đường thẳng)

Ta có :

\(\frac{a.\left(a-1\right)}{2}-1=20\)

\(\frac{a.\left(a-1\right)}{2}=20+1=21\)

\(a.\left(a-1\right)=21.2=47=7.6\)

Vậy a = 7

\(A=\frac{5}{2.4.6}+\frac{5}{4.6.8}+...+\frac{5}{16.18.20}\)

\(4A=5.\left(\frac{4}{2.4.6}+\frac{4}{4.6.8}+...+\frac{4}{16.18.20}\right)\)

\(4A=5.\left(\frac{1}{2.4}-\frac{1}{4.6}+\frac{1}{4.6}-\frac{1}{6.8}+...+\frac{1}{16.18}-\frac{1}{18.20}\right)\)

\(4A=5.\left(\frac{1}{8}-\frac{1}{360}\right)\)

\(4A=5.\left(\frac{45}{360}-\frac{1}{360}\right)\)

\(4A=5.\frac{44}{360}\)

\(A=\frac{5.44}{360.4}\)

\(A=\frac{11}{72}\)

Tham khảo~

\(A=\frac{5}{2\cdot4\cdot6}+\frac{5}{4\cdot6\cdot8}+.....+\frac{5}{16\cdot18\cdot20}\)

\(\frac{4}{5}A=\frac{4}{2\cdot4\cdot6}+\frac{4}{4\cdot6\cdot8}+.........+\frac{4}{16\cdot18\cdot20}\)

\(\frac{4}{5}A=\frac{1}{2\cdot4}-\frac{1}{4\cdot6}+\frac{1}{4\cdot6}-\frac{1}{6\cdot8}+.........+\frac{1}{16\cdot18}-\frac{1}{18\cdot20}\)

\(\frac{4}{5}A=\frac{1}{2\cdot4}-\frac{1}{18\cdot20}\)

\(\frac{4}{5}A=\frac{1}{8}-\frac{1}{360}\)

\(\frac{4}{5}A=\frac{11}{90}\)

\(A=\frac{11}{90}:\frac{4}{5}=\frac{11}{72}\)

\(A=\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+.........+\frac{1}{96\cdot98}\right)-\left(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{97\cdot99}\right)\)

\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+....+\frac{2}{96\cdot98}\right)-\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+.....+\frac{2}{97\cdot99}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{96}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{98}\right)-\frac{1}{2}\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=\frac{12}{49}-\frac{16}{99}=\frac{404}{4851}\)

sai đề nhé?!

Không tồn tại nghiệm số thực.

\([\chi]\in\varnothing\)

ta có

ab-ba=10a+b-10b-a=9a+9b=9.(a+b) chia hết cho 9

vậy ab-ba chia hết cho 9 =>đpcm

ab -ba = 10a +b -(10b+a)

          = 10a +b -10b -a

          = 9a -9b chia hết cho 9

Vậy ab-ba chia hết cho 9