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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
\(1-\frac{1}{5x+6}=\frac{2005}{2006}\Leftrightarrow5x+6=2006\Leftrightarrow x=400\)
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).5x+6}=\frac{2005}{2006}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(1-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\frac{1}{5x+6}=1-\frac{2005}{2006}\)
\(\frac{1}{5x+6}=\frac{1}{2006}\)
=>5x+6 = 2006
=>5x = 2000
=>x = 400
\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
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
\(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\frac{1}{1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(-\frac{1}{5x+6}=\frac{2005}{2006}-\frac{1}{1}\)
\(-\frac{1}{5x+6}=-\frac{1}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2006}\)
⇒ 5x + 6 = 2006
⇒ 5x = 2006 - 6 = 2000
⇒ x = 2000 : 5 = 400
Vậy x = 400