mình làm câu 4 nha

Gọi d là ước chung của 2n+1 và 3n+2 (d thuộc N*)

=>(2n+1) : d và (3n+2) : d

=>3.(2n+1) :d và 2.(3n+2): d

=>(6n+3) :d và (6n+4) : d

=> ((6n+4) - (6n+3)) : d

=>1 :d => d=1

Vì d là ước chung của 2n+1/3n+2

mà d =1 => ƯC(2n+1/3n+2) =1

Vậy 2n+1/3n+2 là phân số tối giản

Tick mình nha bạn hiền .

Đặt \(A=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+3n\left(n+1\right)\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left(n+2-n+1\right)\)

\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)

\(=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Bạn ơi tại sao 3n.(n+1) lại bằng với n.(n+1).(n+2-n+1)

Em xem lại đề câu B nhé\(B=\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+\dfrac{3}{20}+...+\dfrac{3}{\left(n-1\right).n}\\ =3.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{\left(n-1\right).n}\right)\\ =3.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)=3.\left(1-\dfrac{1}{n}\right)=3.\dfrac{n-1}{n}=3-\dfrac{3}{n}.\)

\(C=\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{30.32}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{30}-\dfrac{1}{32}\\ =1-\dfrac{1}{32}=\dfrac{31}{32}.\)

\(D=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n+1}-\dfrac{1}{n+3}\right)\\ =\dfrac{1}{2}.\left(1-\dfrac{1}{n+3}\right)=\dfrac{1}{2}.\dfrac{n+2}{n+3}.\)

C = \(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)

C = \(2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)

C = \(2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\frac{1}{18}\)

C = \(\frac{1}{9}\)

\(B=\frac{6}{1.3}+\frac{6}{3.5}+\frac{6}{5.7}+\frac{6}{7.9}+...+\frac{6}{99.101}\)

\(=3.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{9}{99.101}\right)\)

\(=3.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{99}-\frac{1}{101}\right)\)

\(=3.\left(\frac{1}{1}-\frac{1}{101}\right)=3.\left(\frac{101}{101}-\frac{1}{101}\right)=3.\frac{100}{101}=\frac{300}{101}\)

\(C=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)

\(=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+....+\frac{1}{87}-\frac{1}{90}\right)\)

\(=2.\left(\frac{1}{15}-\frac{1}{90}\right)=2.\left(\frac{6}{90}-\frac{1}{90}\right)=2.\frac{5}{90}=\frac{1}{9}\)

\(\dfrac{6}{1.3}+\dfrac{6}{3.5}+...+\dfrac{6}{99.100}\\ =3\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.100}\right)\\ =3\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =3\left(1-\dfrac{1}{100}\right)\\ =3.\dfrac{99}{100}\\ =\dfrac{297}{100}\)

bn lên ngạng hoặc và xem câu hỏi tương tự nha!

Nhớ k mk đấy nha!

thanks nhìu!

OK..OK..OK

\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(2C=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)

Ta có : 

\(\frac{2}{1.3}=1-\frac{1}{3}\)

\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)

...............................

\(\frac{2}{\left(2n-1\right)\left(2n+1\right)}=\frac{1}{2n-1}-\frac{1}{2n+1}\)

\(\Rightarrow2C=1-\frac{1}{2n+1}=\frac{2n}{2n+1}\)

\(\Rightarrow C=\frac{n}{2n+1}\)

BẠN CÓ THỂ VÀO CÂU HỎI TƯƠNG TỰ .