Tìm x :
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+..+\frac{1}{\left(5x+1\right)\left(5x+6\right)}=\frac{10}{41}\)
tìm x
d) \(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2005}{2006}\)
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
\(1-\frac{1}{5x+6}=\frac{2005}{2006}\Leftrightarrow5x+6=2006\Leftrightarrow x=400\)
tìm x,y thỏa mãn
a.\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)
\(\Rightarrow5x+6=2011\)
\(\Rightarrow5x=2011-6\)
\(\Rightarrow5x=2005\)
\(\Rightarrow x=401\)
\(\frac{5}{1.6}+\frac{5}{6.11}+..+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)
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
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2005}{2006}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(1-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\frac{1}{5x+6}=1-\frac{2005}{2006}\)
\(\frac{1}{5x+6}=\frac{1}{2006}\)
\(\Rightarrow5x+6=2006\)
\(5x=2006-6\)
\(5x=2000\)
\(x=2000:5\)
\(x=400\)
CMR: mọi n thuộc N ta có
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5v+1\right).\left(5n+6\right)}=\frac{n+1}{5n+6}\)
Tính:
\(D=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right).\left(5n+6\right)}\)
Tính đầy đủ hộ mik vs. Mik đag cần gấp
D = \(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}\)
= \(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{5n+1}-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}\left(1-\frac{1}{5n+6}\right)\)
= \(\frac{1}{5}.\frac{5n+5}{5n+6}\)
= \(\frac{n+1}{5n+6}\)
Chứng minh rằng với mọi n \(\in\) N ta luôn có:
\(\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{\left(5n+1\right)\left(5n+6\right)}=\frac{n+1}{5n+6}\)
Heo mi pờ lít
câu hỏi tương tự có đó bạn, bạn vào tham khảo nhe!
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{10000}\right)\)
\(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
Tính B và D
\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{10000}\right)\)
\(=\left(\frac{4}{4}-\frac{1}{4}\right).\left(\frac{9}{9}-\frac{1}{9}\right)...\left(\frac{10000}{10000}-\frac{1}{10000}\right)\)
\(=\frac{3}{4}.\frac{8}{9}...\frac{9999}{10000}=\frac{3}{2.2}.\frac{2.4}{3.3}...\frac{99.101}{100.100}\)
\(=\frac{101}{100}\)
\(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\)
\(=5.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=5.\left(\frac{1}{1}-\frac{1}{31}\right)=5.\left(\frac{31}{31}-\frac{1}{31}\right)=5.\frac{30}{31}=\frac{150}{31}\)
Tìm x, biết:
\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
Các bạn giải cụ thể giùm mình nha
\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
chứng minh rằng với mọi n thuộc Z ta luôn \(\frac{1}{1.6}\)+ \(\frac{1}{6.11}\)+\(\frac{1}{11.16}\)+........+\(\frac{1}{\left(5n+1\right).\left(5n+6\right)}\)=\(\frac{n+1}{5n+6}\)
giúp mình đi sớm nhé