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

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

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!~

 ta có A =\(\frac{1}{5\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot15}+...+\frac{1}{605\cdot608}\)

3A =\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{605\cdot608}\)

3A =\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\)

3A=\(\frac{1}{5}-\frac{1}{608}\)

3A=\(\frac{603}{3040}\)A =\(\frac{201}{3040}\)

Đặt A=\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\)

      3A=\(3.\left(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{605.608}\right)\)

      3A=\(3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{605}-\frac{1}{608}\right)\)

      3A=3.\(\left(\frac{1}{5}-\frac{1}{608}\right)\)

       A=\(\frac{201}{3040}\)

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 ???

\(3S=3\left(\frac{1}{2.5}+....+\frac{1}{\left(3n+1\right)\left(3n+2\right)}\right)\)

Đến đây thì bạn làm như dạng đơn giản nhé

 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)1/5.8+1/8.11+1/11.14+...+1/x(x+3)=101/1540

<=>1/3(3/5.8+3/8.11+...+3/x(x+3)     =101/1540

<=>1/3(1/5-1/8+1/8-1/11+...+1/x-1/x+3=101/1540

<=>1/5-1/x+3=303/1540<=>1/x+3=1/308

<=>x+3=308<=>x=305

Nguồn CHTT, hihi !

Tham gia event này đi mọi người https://olm.vn/hoi-dap/detail/227766827875.html

x=1

duyệt đi

\(\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{65.68}\right)x=\frac{19}{68}+\frac{7}{34}\)

\(\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{68}\right)x=\frac{33}{68}\)

\(\left(\frac{1}{2}-\frac{1}{68}\right)x=\frac{33}{68}\)

\(\frac{33}{68}x=\frac{33}{68}\)

\(x=\frac{33}{68}:\frac{33}{68}=1\)

x=3

nhoc quay phá làm thiếu 

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

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