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\(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}=\frac{31}{1.3}+\frac{31}{3.5}+\frac{31}{5.7}+\frac{31}{7.9}+\frac{31}{9.11}+\frac{31}{11.13}\\ \)
\(=\frac{31}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{31}{2}.\left(1-\frac{1}{13}\right)=\frac{31}{2}.\frac{12}{13}=\frac{31.6}{13}=\frac{186}{13}\)
\(\Rightarrow x-\frac{186}{13}=\frac{9}{13}\Leftrightarrow x=\frac{195}{13}=15\)
\(x-\left(\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\right)=\)\(\frac{9}{13}\)(1)
Đặt \(A=\frac{31}{3}+\frac{31}{15}+\frac{31}{35}+\frac{31}{63}+\frac{31}{99}+\frac{31}{143}\)
\(A=31\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\right)\)
\(\Rightarrow2A=31\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)\)
\(2A=31\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}\right)\)
\(2A=31\left(2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(2A=31\left(2-\frac{1}{13}\right)\)
\(2A=31.\frac{25}{13}\)
\(2A=\frac{775}{13}\)
\(\Rightarrow A=\frac{775}{13}:2\)
\(A=\frac{775}{26}\)
Thay vào (1) ta có:
\(x-\frac{775}{26}=\frac{9}{13}\)
\(\Leftrightarrow x=\frac{9}{13}+\frac{775}{26}\)
\(\Leftrightarrow x=\frac{61}{2}\)
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[\left(35\frac{5}{7}-5\frac{5}{7}\right)+2\frac{3}{4}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[30+3\right]:\left(11+x\right)=3\)
\(33:\left(11+x\right)=3\)
\(11+x=11\)
\(x=0\)
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{35\times7+5}{7}+\frac{2\times4+3}{4}\right)-\frac{5\times7+5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{250}{7}+\frac{11}{4}\right)-\frac{40}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\frac{1077}{28}-\frac{167}{28}\right]\div\left(11+x\right)=3\)
\(32,5\div\left(11+x\right)=3\)
\(11+x=32,5\div3\)
\(11+x=\frac{65}{6}\)
\(x=\frac{65}{6}-11=-\frac{1}{6}\)
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
Vậy X=109/6075
Chắc Sai kết quả chứ công thức đúng nha!!!...
Fighting!!!...
Đặt:
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{25-23}{23.25}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Đặt \(B=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\right)=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(2B=\frac{242}{243}\Rightarrow B=\frac{121}{243}\)
Giải phương trình:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
Nhân 2 cả 2 vế lên:
\(\left(2x+\frac{2}{1x3}\right)+...+\left(2x+\frac{2}{23x25}\right)=22x+\frac{2}{3}+\frac{2}{9}+\frac{2}{81}+\frac{2}{243}\)2/243
\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)
\(24x+\frac{24}{25}=22x+\frac{224}{243}\)
\(2x=\frac{224}{243}-\frac{24}{25}\)
\(2x=-\frac{232}{6025}\)
\(x=\frac{-116}{6075}\)
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)
\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)
\(12x+\frac{12}{25}=11x+\frac{112}{243}\)
\(11x-12x=\frac{112}{243}-\frac{12}{25}\)
\(-1x=-\frac{116}{6075}\)
\(x=-\frac{116}{6075}\div\left(-1\right)\)
\(x=\frac{116}{6075}\)
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
bạn Kudo Shinichi sai rồi: muốn tìm số chia mà thương chia cho số bị chia à
a) 7,2 * x = 6,49
x = 6,49 : 7,2
x = 649/720
b) \(\frac{15}{77}:x=\frac{3}{11}\)
\(x=\frac{3}{11}:\frac{15}{77}=\frac{7}{5}\)
c) 2,4 * x + 1,1 *x = 0,7
x * ( 2,4 + 1,1 ) = 0,7
x * 3,5 = 0,7
=> x = 0,7 : 3,5 = 0,2
phần d sai đề đó bạn, 25/31 chia rồi lại công là sai, sửa đề đi nha.
Ai thấy đúng thì ủng hộ nha !!!