ta có: \(S=1+1\times2+2\times3+3\times4+...+38\times39+39\times40+40\)

\(\Rightarrow3S=1\times3+1\times2\times3+2\times3\times3+...+39\times40\times3+40\times3\)

\(3S=3+1\times2\times\left(3-0\right)+2\times3\times\left(4-1\right)+...+39\times40\times\left(41-38\right)+120\)

\(3S=3+1\times2\times3+2\times3\times4-1\times2\times3+...+39\times40\times41-38\times39\times40+120\)

\(3S=\left(3+1.2.3+...+39.40.41+120\right)-\left(1.2.3+...+38.38.40\right)\)

\(3S=3+39.40.41+120\)

\(\Rightarrow S=\left(3+39.40.41+120\right):3\)

\(S=21361\)

1-1=0

2-1=3

3-1=8

4-1=15

4-1=24

Why ??? giải thích ra , mà thôi mk bt KQ rùi , thôi đành k cho  bn zậy

Bài 1 \(F=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)

        \(2F=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{7.8}-\frac{1}{8.9}+\frac{1}{8.9}-\frac{1}{9.10}\)

        \(2F=\frac{1}{1.2}-\frac{1}{9.10}\)\(=\frac{44}{90}\)

          \(F=\frac{11}{45}\)

Vậy \(F=\frac{11}{45}\)

Bài 2 : 

\(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)

    \(\Rightarrow\)\(\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{9.9}\)

    \(\Rightarrow\)\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}< B< \frac{1}{2.3}+..+\frac{1}{8.9}\)

     \(\Rightarrow\)\(\frac{1}{3}-\frac{1}{10}< B< \frac{1}{2}-\frac{1}{9}\)

     \(\Rightarrow\)\(\frac{7}{30}\)\(< \frac{7}{18}\left(đpcm\right)\)

Hết nha bn.Mk ik ngủ.Chúc bạn học tốt

Vai trò a,b như sau: 

Gỉa sử \(a\ge b\Rightarrow a=b=m\left(m\ge0\right)\)

\(\frac{a}{b}+\frac{b}{a}=\frac{b+m}{b}+\frac{b}{b+m}\)

                \(=\)\(\frac{b}{b}+\frac{m}{b}+\frac{\left(b+m\right)-m}{b+m}\)

                  \(=1+\frac{m}{b}+1-\frac{m}{b+m}\)

Vì \(\frac{m}{b}\ge\frac{m}{b+m}\)

\(\Rightarrow\)\(\frac{m}{b}-\frac{m}{b+m}\ge0\)

Vậy \(\frac{a}{b}+\frac{b}{a}\ge2\left(Đpcm\right)\)

Bạn gửi nhanh nha.Chúc bạn học tốt

Ta có: \(25\equiv6\)\(\)(mod 6)

\(\Rightarrow\)\(25^n\equiv6^n\)(mod 19)

\(A=7.25^n+12.6^n\)

\(A=7.6^n+12.6^n\)

\(A=6^n.\left(7+12\right)\)

\(A=6^n.19\equiv0\)(mod19)

Vậy  \(A⋮19\left(đpcm\right)\)

Bạn học tốt nha !Mk hứa vs bn trả lời 5  câu.Mak cn bn gửi ik

ta có: \(A=7.25^n+12.6^n\)

\(A=7.25^n-7.6^n+19.6^n\)

\(A=7.\left(25^n-6^n\right)+19.6^n\)

mà \(25^n-6^n⋮25-6\)

\(\Rightarrow25^n-6^n⋮19\)

\(\Rightarrow7.\left(25^n-6^n\right)⋮19\)

 mà \(19.6^n⋮19\)

\(\Rightarrow7.\left(25^n-6^n\right)+19.6^n⋮19\)

\(\Rightarrow A⋮19\left(đpcm\right)\)

0,2.17.8+0,16.540+29.1,6=(0,2.17.8)+(0,16.540)+(29.1,6)=160

\(0,2.17,8+0,16.540+29.1,6\)

\(=27,2+86,4+46,4\)

\(=27,2+132,8\)

\(=160\)

_Chúc bạn học tốt_

Ta có : \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

           \(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2015.2017}\right)\)

           \(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

            \(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

             \(=\frac{1}{2}.\frac{2016}{2017}=\frac{1008}{2017}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2015.2017}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(=\frac{1}{2}.\frac{2016}{2017}\)

\(=\frac{1008}{2017}\)

D/ listen to

D/ listen to

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