\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+\frac{3}{14\cdot17}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

Tính

\(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+\frac{3}{14\times17}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)

\(=\frac{1}{2}-\frac{1}{17}=\frac{17}{34}-\frac{2}{34}=\frac{15}{34}\)

Ta có: Biểu thức trên sẽ =

\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}\)1/17 (nó bị lỗi nên k viết đc T_T)

= \(\frac{1}{2}-\frac{1}{17}=\frac{15}{34}\)

Chúc bạn học tốt!~

Đề là cm S>1 nha bạn!

\(S=\frac{9}{2.5}+\frac{9}{5.8}+...+\frac{9}{29.32}\)

\(=3\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{29.32}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{29}-\frac{1}{32}\right)\)

\(=3\left(\frac{1}{2}-\frac{1}{32}\right)\)

\(=3.\frac{15}{32}\)

\(=\frac{45}{32}>1\)

\(\Leftrightarrow S>1\)

\(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}\)

Cách 1 : Vì hiệu hai thừa số đều là 3 = 5 - 2 = 8 - 5 = ... = 32 - 29 nên phân tích tử 9 = 3 . 3

Ta có : \(S=3\left[\frac{3}{2\cdot5}+\frac{3}{7\cdot9}+...+\frac{3}{29\cdot32}\right]=3\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{29}-\frac{1}{32}\right]\)

\(=3\left[\frac{1}{2}-\frac{1}{32}\right]=3\left[\frac{16}{32}-\frac{1}{32}\right]=3\cdot\frac{15}{32}=\frac{45}{32}\)

Mà \(\frac{45}{32}>1\)=> S không thể bé hơn 1

Cách 2 : Nhận xét : \(\frac{9}{2\cdot5}=\frac{3}{2}-\frac{3}{5};\frac{9}{5\cdot8}=\frac{3}{5}-\frac{3}{8};...\)

Vậy ta có : \(S=\frac{9}{2\cdot5}+\frac{9}{5\cdot8}+\frac{9}{8\cdot11}+...+\frac{9}{29\cdot32}=\frac{3}{2}-\frac{3}{5}+\frac{3}{5}-\frac{3}{8}+...+\frac{3}{29}-\frac{3}{32}\)

\(=\frac{3}{2}-\frac{3}{32}=\frac{3\cdot16}{32}-\frac{3}{32}=\frac{48}{32}-\frac{3}{32}=\frac{45}{32}\)

Tự so sánh , mà S đâu bé hơn 1 ???

a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right]\)

\(=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3n+2}\right]=\frac{1}{3}\left[\frac{3n+2}{2\left(3n+2\right)}-\frac{2}{2\left(3n+2\right)}\right]\)

\(=\frac{1}{3}\cdot\frac{3n}{6n+4}=\frac{n}{6n+4}=VP\)

b) Ta có: \(\frac{5}{3.7}+\frac{5}{7.11}+...+\frac{5}{\left(4n-1\right)\left(4n+3\right)}\)

\(=\frac{5}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{\left(4n-1\right)\left(4n+3\right)}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{4n-1}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{1}{3}-\frac{1}{4n+3}\right)\)

\(=\frac{5}{4}\left(\frac{4n+3}{12n+9}-\frac{3}{12n+9}\right)\)

\(=\frac{5}{4}.\frac{4n}{12n+9}\)

\(=\frac{5n}{12n+9}\)

( sai đề )

\(\frac{16}{11},-\frac{5}{9},\frac{10}{539}\)

 bÀI LÀM

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

A=2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

A= 2 - 1/3 + 1/3 - 1/5 + 1/5 - ... + 2/99 - 2/101

A = 2 - 2/101 = 200/101

B = 3-1/3+1/3-1/5+1/5-...+3/49-3/51

B = 3-3/51(tự tính nhé)

C = 5(5/1.6+5/6.11+5/11.16+....+5/26-5/31

C = 5(5-1/31)(tự tính)

D rút gon cho 2 rồi 3D , sau đó 5(3/.... tương tự các cách làm trên)

2E nhân lên rồi giải giống trên

3F Rồi nhân 4/77 và rút gọn thì tính được

a, A= \(\frac{1}{1}\)- \(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+......+\(\frac{1}{99}\)-\(\frac{1}{100}\)

A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+(-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....-\(\frac{1}{99}\)+\(\frac{1}{99}\))

A=\(\frac{1}{1}\)-\(\frac{1}{100}\)+0

A=1-\(\frac{1}{100}\)=\(\frac{100}{100}\)-\(\frac{1}{100}\)=\(\frac{99}{100}\)

a) A= \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\) 

=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

=\(1-\frac{1}{100}=\frac{99}{100}\)

b) \(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}\)

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

=\(\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right).\frac{3}{2}\)

= \(\left(1-\frac{1}{50}\right).\frac{3}{2}=\frac{49}{50}.\frac{3}{2}=\frac{147}{100}\)

c) \(C=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{26.31}\)

= \(\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{26.31}\right).5\)

= \(\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{26}-\frac{1}{31}\right).5\) 

= \(\left(1-\frac{1}{31}\right).5=\frac{30}{31}.5=\frac{150}{31}\)

Mấy bài còn lại mik đang phải nháp đã. Bạn thông cảm cho mik

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\)\(\frac{3}{11\cdot14}+...+\)\(\frac{3}{602\cdot605}\)

\(=\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{602\cdot605}\)

\(=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}\)\(-\frac{1}{605}\)

\(=\frac{1}{5}-\frac{1}{605}\)

\(=\frac{121}{605}-\frac{1}{605}\)

\(=\frac{120}{605}=\frac{24}{121}\)

Bài này dùng công thức nhé 

\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{602.605}\)

\(=\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\)\(\frac{1}{5}-\frac{1}{605}\)

\(=\)\(\frac{24}{121}\)

Chúc bạn học tốt ~ 

\(A=\frac{5-2}{2x5}+\frac{8-5}{5x8}+\frac{11-8}{8x11}+...+\frac{20-17}{17x20}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{17}-\frac{1}{20}\)

\(A=\frac{1}{2}-\frac{1}{20}=\frac{9}{20}\)