em gửi bài qua fb của thầy thầy HD nhé: tìm fb của thầy bằng sđt:0975705122 nhé

k mình nha

Câu 1:

\(-\frac{1}{54}-\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{79.81}\right)\)

\(=-\frac{1}{54}-\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{79}-\frac{1}{81}\right)\)

\(=-\frac{1}{54}-\frac{3}{2}\left(1-\frac{1}{81}\right)\)

\(=-\frac{1}{54}-\frac{40}{27}\)

\(=-\frac{3}{2}\)

Câu 2:

\(a^2+b^2+c^2+d^2+e^2=\left(a+b+c+d+e\right)^2-2\left(ab+ac+ad+ae+bc+bd+be+cd+ce+de\right)\)

Mà \(2\left(ab+ac+ad+ae+bc+bd+be+cd+ce+de\right)⋮2\)

\(\Rightarrow\left(a+b+c+d+e\right)^2⋮2\)

\(\Rightarrow a+b+c+d+e⋮2\)

Do \(a,b,c,d,e\) nguyên dương \(\Rightarrow a+b+c+d+e>2\Rightarrow a+b+c+d+e\) là hợp số

Câu 3:

- Chiều thuận: \(3a+2b⋮17\Rightarrow10a+b⋮17\)

Ta có \(\left\{{}\begin{matrix}17a⋮17\\3a+2b⋮17\end{matrix}\right.\) \(\Rightarrow17a+3a+2b⋮17\Rightarrow20a+2b⋮17\)

\(\Rightarrow2\left(10a+b\right)⋮17\), mà 2 và 17 nguyên tố cùng nhau \(\Rightarrow10a+b⋮17\)

- Chiều nghịch: \(10a+b⋮17\Rightarrow3a+2b⋮17\)

\(10a+b⋮17\Rightarrow2\left(10a+b\right)⋮17\Rightarrow20a+2b⋮17\)

\(\Rightarrow17a+3a+2b⋮17\)

Mà \(17a⋮17\Rightarrow3a+2b⋮17\) (đpcm)

\(C=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-...-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

\(C=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\right)\)

\(C=\frac{3}{2}\left(1-\frac{1}{101}\right)\)

\(C=\frac{3}{2}.\frac{100}{101}=\frac{150}{101}\)

C = 31.3+33.5+35.7+...+399.101

C = 3\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

C =3\(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

C =3\(\left(1-\frac{1}{101}\right)\)

C=3 . \(\frac{100}{101}\)

C=\(\frac{300}{101}\)

Ta có: B = 2²⁰¹⁰  -   2²⁰⁰⁹  -  2²⁰⁰⁸  -......- 2 -1

=> B = 2²⁰¹⁰ - (1 + 2 + 2² + ..... + 2²⁰⁰⁹)

Đặt A = 1 + 2 + 2² + .... + 2²⁰⁰⁹

=> 2A = 2 + 2² + .... + 2²⁰¹⁰

=> 2A - A = 2²⁰¹⁰ - 1

=> A = 2²⁰¹⁰ - 1

Vậy B = 2²⁰¹⁰ - (2²⁰¹⁰ - 1)

=> B = 2²⁰¹⁰ - 2²⁰¹⁰  + 1

=> B = 1

Ta có: B = 2²⁰¹⁰  -   2²⁰⁰⁹  -  2²⁰⁰⁸  -......- 2 -1

=> B = 2²⁰¹⁰ - (1 + 2 + 2² + ..... + 2²⁰⁰⁹)

Đặt A = 1 + 2 + 2² + .... + 2²⁰⁰⁹

=> 2A = 2 + 2² + .... + 2²⁰¹⁰

=> 2A - A = 2²⁰¹⁰ - 1

=> A = 2²⁰¹⁰ - 1

Vậy B = 2²⁰¹⁰ - (2²⁰¹⁰ - 1)

=> B = 2²⁰¹⁰ - 2²⁰¹⁰  + 1

=> B = 1

Ta có: B = 2²⁰¹⁰  -   2²⁰⁰⁹  -  2²⁰⁰⁸  -......- 2 -1

=> B = 2²⁰¹⁰ - (1 + 2 + 2² + ..... + 2²⁰⁰⁹)

Đặt A = 1 + 2 + 2² + .... + 2²⁰⁰⁹

=> 2A = 2 + 2² + .... + 2²⁰¹⁰

=> 2A - A = 2²⁰¹⁰ - 1

=> A = 2²⁰¹⁰ - 1

Vậy B = 2²⁰¹⁰ - (2²⁰¹⁰ - 1)

=> B = 2²⁰¹⁰ - 2²⁰¹⁰  + 1

=> B = 1

=3(1/1.3+1/3.5+1/5.7+1/7.9)

=3/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) vi khoang cach tu 1-3;3-5;5-7;7-9 la 2 nen ta nhan tat ca voi 1/2 ma 3.1/2=3/2

=3/2.(1-1/9) rut gon -1/3+1/3;-1/5+1/5;-1/7+1/7=0

=3/2.8/9=4/3

ta có :3/(1.3)+3/(3.5)+3/(5.7)+3/(7.9)

ta đặt 3 làm chung rồi tự làm đc

dat 3 lam chung

C=\(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

C=\(\frac{1}{6}\left\{\frac{6}{1.7}+\frac{6}{7.13}+\frac{6}{13.19}+...+\frac{6}{31.37}\right\}\)=\(\frac{1}{6}\left(1-\frac{1}{7}+\frac{1}{7}+....+\frac{1}{31}-\frac{1}{37}\right)\)

C=\(\frac{1}{6}\left(1-\frac{1}{37}\right)=\frac{1}{6}.\frac{36}{37}=\frac{36}{222}=\frac{6}{37}\)

D=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+......+\frac{3}{49.51}\)

D=\(\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{49.51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

D=\(\frac{3}{2}\left(1-\frac{1}{51}\right)=\frac{3}{2}.\frac{50}{51}\)

D=\(\frac{150}{102}\)=\(\frac{25}{17}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)

\(A=1-\frac{1}{51}\)

\(A=\frac{50}{51}\)

\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)

\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)

\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(2A=3\left(1-\frac{1}{51}\right)\)

\(2A=3.\frac{50}{51}\)

\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'

A=3/2(2/1.3+2/3.5+2/5.7+....+2/53.55)

=3/2(1-1/3+1/3-1/5+1/5-1/7+..../1/53-1/55)

=3/2(1-1/55)

=3/2.54/55

=81/55