Câu 1 : Chú ý chữ x là dấu ích , ko phải nhân đâu nha.Phép tính liên quan đến phân số. Một dấu chấm là nhân . Hai phép đầu là tìm x
1/3 + 5/6 : ( x - 2 và 1/5 ) = 3/4
1/3 + 1/6 + 1/10 + ....... + 1/ x . ( x + 1 ): 2 = 2009 / 2011
1/15 + 1/35 + 1/63 + 1/99 + ........... + 1/2915 + 1/3135 = ?
AI làm được phép nào thì làm giúp mk ,mk ko bắt buộc đâu.
a) \(\frac{1}{3}+\frac{5}{6}:\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)
=> \(\frac{1}{3}+\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{3}{4}-\frac{1}{3}\)
=> \(\frac{5}{6}:\left(x-\frac{11}{5}\right)=\frac{5}{12}\)
=> \(x-\frac{11}{5}=\frac{5}{6}:\frac{5}{12}\)
=> \(x-\frac{11}{5}=2\)
=> \(x=2+\frac{11}{5}\)
=> \(x=\frac{21}{5}\)
1/15 +1/35 + 1/63 + ... + 1/3135
= 1/3*5 + 1/5*7 + 1/7*9 + ... + 1/55*57
= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/55 - 1/57
= 1/3 - 1/57
= 18/57
1/3 + 1/6 + 1/10 + ... + 1/x(x + 1)*2 = 2009/2011
=> 2/6 + 2/12 + 2/20 + ... + 2/x(x + 1)*2 = 2009/2011
=> 2(1/6 + 1/12 + 1/20 + ... + 1/x(x + 1)) = 2009/2011
=> 2(1/2*3 + 1/3*4 + 1/4*5 + ... + 1/x(x + 1)) = 2009/2011
=> 2(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x + 1) = 2009/2011
=> 2(1/2 - 1/x + 1) = 2009/2011
=> 1/2 - 1/x + 1 = 2009/4022
=> 1/x + 1 = 1/2011
=> x + 1 = 2011
=> x = 2010
vậy-
#)Giải :
a)\(\frac{1}{3}+\frac{5}{6}\div\left(x-2\frac{1}{5}\right)=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{3}+\frac{5}{6}\div\left(x-\frac{11}{5}\right)=\frac{3}{4}\)
\(\Leftrightarrow\frac{5}{6}\div\left(x-\frac{11}{5}\right)=\frac{5}{12}\)
\(\Leftrightarrow x-\frac{11}{5}=2\)
\(\Leftrightarrow x=\frac{21}{5}\)
b)Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{x\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(\Leftrightarrow A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow A=2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow A=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow A=1-\frac{2}{x+1}=\frac{2009}{2011}\)
Đến đây thì ez rùi nhé ^^
b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{\frac{1}{x.\left(x+1\right)}}{2}=\frac{2009}{2011}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+...+\frac{\frac{1}{2}.\frac{1}{2}}{2}\right)=\frac{2009}{2011}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+...+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
=> \(\frac{1}{x+1}=\frac{1}{2011}\)
Vì 1 = 1
=> x + 1 = 2011
=> x = 2011 - 1
=> x = 2010
