Ta có công thức \(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)

Dựa vào công thức trên, ta có:

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Rightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Rightarrow\)\(\frac{1}{5x+6}=1-\frac{2005}{2006}=\frac{1}{2006}\)

\(\Rightarrow\)\(5x+6=2006\Rightarrow x=400\)

chắc chắn, ủng hộ mink nha

         \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(1-\frac{1}{5x+6}=\frac{2005}{2006}\)

        \(\frac{1}{5x+6}=1-\frac{2005}{2006}\)

        \(\frac{1}{5x+6}=\frac{1}{2006}\)

\(\Rightarrow5x+6=2006\)

             \(5x=2006-6\)

            \(5x=2000\)

               \(x=2000:5\)

               \(x=400\)

1-1/6+1/6-1/11+...+1/5x+1-1/5x+6=2005/2006

1-1/5x+6=1-1/2006

5x+6=2006

5x=2000

x=400

\(1-\frac{1}{5x+6}=\frac{2005}{2006}\Leftrightarrow5x+6=2006\Leftrightarrow x=400\)

x=400 

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)

\(\Rightarrow5x+6=2011\)

\(\Rightarrow5x=2011-6\)

\(\Rightarrow5x=2005\)

\(\Rightarrow x=401\)

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).5x+6}=\frac{2005}{2006}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\frac{1}{5x+6}=1-\frac{2005}{2006}\)

\(\frac{1}{5x+6}=\frac{1}{2006}\)

=>5x+6 = 2006

=>5x = 2000

=>x = 400

x = 400 nha bạn

                       Bài làm :

Ta có :

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)

\(\Leftrightarrow\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{5x+6-5x-1}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)

\(\Leftrightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Leftrightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)

\(\Leftrightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)

\(\Leftrightarrow\frac{1}{5x+6}=\frac{1}{2006}\)

\(\Leftrightarrow1.2006=\left(5x+6\right).1\)

\(\Leftrightarrow2006=5x+6\)

\(\Leftrightarrow5x=2006-6\)

\(\Leftrightarrow5x=2000\)

\(\Leftrightarrow x=2000\div5=400\)

Vậy x=400

Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)

\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)

\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)

\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)

\(\frac{20}{21}x=\frac{1}{25}\)

\(x=\frac{1}{25}:\frac{20}{21}=.....\)

không ai trả lời đâu

a,\(\frac{x+1}{2}\)\(=\frac{8}{x+1}\)

\(\Leftrightarrow\)(x+1)\(\times\)(x+1) = 8 \(\times\)2

\(\Leftrightarrow\)(x+1)² = 16

\(\Leftrightarrow\)(x+1)² = 4²

\(\Rightarrow\)x+1 = 4

\(\Rightarrow\)x = 4 - 1

\(\leftrightarrow\)x = 3

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

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

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

\(A=5.\frac{30}{31}\)

\(A=\frac{150}{31}>1\)

Đề hơi lạ nhỉ, vì quá rõ ràng rùi 5²/1.6 = 25/6 > 1 nên A lớn hơn 1

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

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

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

   \(=5\left(1-\frac{1}{31}\right)\)

    \(=5.\frac{30}{31}=\frac{150}{31}>1\)

=> A > 1

Study well ! >_<