\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{2006.2009}\)

\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(F=\frac{1}{5}-\frac{1}{2009}\)

\(F=\frac{2004}{10045}\)

\(F=\frac{3}{5.8}+\frac{3}{8.11}+\frac{1}{11.14}+...+\frac{3}{2006.2009}\)

\(F=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{2006}-\frac{1}{2009}\)

\(F=\frac{1}{5}-\frac{1}{2009}\)

\(F=0\)

F = 1/5 - 1/8 + 1/8 - 1/11 +,,,+1/2006 - 1/2009
F = 1/5 - 1/2009
F = 2008/10044
Chúc bạn học tốt 

\(\frac{3^2}{2\cdot11}+\frac{3^2}{11\cdot14}+...+\frac{3^2}{197\cdot200}=\frac{3^2}{2\cdot11}+\left(\frac{3^2}{11\cdot14}+...+\frac{3^2}{197\cdot200}\right)\)

\(=\frac{9}{22}+3\left(\frac{3}{11\cdot14}+...+\frac{3}{197\cdot200}\right)=\frac{9}{22}+3\left(\frac{1}{11}-\frac{1}{14}+...+\frac{1}{197}-\frac{1}{200}\right)\)

\(=\frac{9}{22}+3\left(\frac{1}{11}-\frac{1}{200}\right)=\frac{9}{22}+3\left(\frac{200}{2200}-\frac{11}{2200}\right)=\frac{9}{22}+3\cdot\frac{189}{2200}\)

\(=3\cdot\left(\frac{3}{22}+\frac{189}{2200}\right)=3\cdot\left(\frac{300}{2200}+\frac{189}{2200}\right)=3\cdot\frac{489}{2200}=\frac{1467}{2200}\)

\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{97.100}\)

\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{97.100}\right)\)

\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}.\frac{49}{100}=\frac{49}{300}\)

Ta có: \(S=\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{97.100}.\)

\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{97.100}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\)

\(\Rightarrow3S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(\Rightarrow S=\frac{49}{100}:3=\frac{49}{300}\)

Vậy \(S=\frac{49}{300}\)

CHÚC BẠN HỌC TỐT

\(S=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\)

\(S=3\cdot\frac{1}{3}\cdot\left(\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+\frac{1}{11\cdot14}+...+\frac{1}{97\cdot100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}+...+\frac{3}{97\cdot100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\left(\frac{50}{100}-\frac{1}{100}\right)\)

\(S=\frac{1}{3}\cdot\frac{49}{100}\)

\(S=\frac{49}{300}\)

a) \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

                                                              \(=1-\frac{1}{32}=\frac{31}{32}\)

b) \(\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}\)\

\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(\frac{1}{4}-\frac{1}{6}=\frac{1}{12}\)

\(B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{23.26}\)

\(B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{23}-\frac{1}{26}\)

\(B=\frac{1}{2}-\frac{1}{26}\)

\(B=\frac{13}{26}-\frac{1}{26}\)

\(B=\frac{12}{26}=\frac{4}{13}\)

Ta có: 3/2.5=3/3.(1/2-1/5)

3/5.8=3/3.(1/5-1/8)

3/8.11=3/3.(1/8-1/11)

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

3/23.26=3/3.(1/23-1/26)

Cộng từng vế ta đc:

B=3/3.(1/5-1/26)=1.21/130=21/130

B = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - ... - 1/26

B = 1/2 - 1/26

B = 6/13 nha bạn

 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

 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=\(\frac{201}{3040}\)