Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\frac{2020.125+1000}{126.2020-1020}=\frac{2020.125+1000}{2020.125+2020-1020}=1\)
b,\(\left(\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{19.21}\right).426+x=19\)
\(< =>\left(\frac{1}{11}-\frac{1}{21}\right).213+x=19\)
\(< =>\frac{2130}{231}+x=19\)
\(< =>x=19-\frac{2130}{231}=...\)
a)\(\frac{2020\cdot125+1000}{126\cdot2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+2020-1020}=\frac{2020\cdot125+1000}{2020\cdot125+1000}=1\)
b) \(\left(\frac{1}{11\cdot13}+\frac{1}{13\cdot15}+\frac{1}{15\cdot17}+\frac{1}{17\cdot19}+\frac{1}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+\frac{2}{17\cdot19}+\frac{2}{19\cdot21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{11}-\frac{1}{21}\right)\cdot426+x=19\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{10}{231}\cdot426+x=19\)
\(\Leftrightarrow\frac{710}{77}+x=19\)
\(\Leftrightarrow x=19-\frac{710}{77}=\frac{753}{77}\)
Sao đang phép trừ thành phép cộng vậy bạn. Nếu cọng hết thì mik bik tính đó.
\(M=\frac{3}{2}.\left(\frac{2}{11.13}+\frac{2}{13.15}+......+\frac{2}{97.99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+.....+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{11}-\frac{1}{99}\right)=\frac{3}{2}.\frac{8}{99}=\frac{4}{33}\)
M= \(\frac{3}{11\cdot13}+\frac{3}{13\cdot15}+\frac{3}{15\cdot17}+...+\frac{3}{97\cdot99}\)
=\(\frac{3}{2}\cdot\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+\frac{2}{15\cdot17}+...+\frac{2}{97\cdot99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\left(\frac{1}{11}-\frac{1}{99}\right)\)
=\(\frac{3}{2}\cdot\frac{8}{99}\)
= \(\frac{4}{33}\)
Gọi dãy trên là A
\(\Leftrightarrow2A=\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{19\cdot21}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\)
\(\Leftrightarrow2A=\frac{1}{11}-\frac{1}{21}+0+...+0\)
\(\Leftrightarrow2A=\frac{10}{231}\)
\(\Leftrightarrow A=\frac{5}{231}\)
Đặt biểu thức đó là A
=> 2A = \(\frac{2}{9.11}+\frac{2}{11.13}+...+\frac{2}{61.63}=\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{61}-\frac{1}{63}\)
\(=\frac{1}{9}-\frac{1}{63}=\frac{2}{21}\)
=> A = \(\frac{2}{21}.\frac{1}{2}=\frac{1}{21}\)
\(A=\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{54.55}\)
\(A=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{54}-\frac{1}{55}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{11}-\frac{1}{55}\right)\)
\(A=\frac{1}{2}.\frac{4}{55}\)
\(A=\frac{2}{55}\)