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\(\frac{1}{4}:\frac{x}{9}=\frac{1}{4}\)
\(\frac{x}{9}=\frac{1}{4}:\frac{1}{4}\)
\(\frac{x}{9}=1\)
\(\Rightarrow x=9\)
\(\left(X+\frac{1}{1.3}\right)+\left(X+\frac{1}{3.5}\right)+...+\left(X+\frac{1}{23.25}\right)=11.X+\)\(\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(\Leftrightarrow12X+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)+11X\)\(+\frac{\left(1+\frac{1}{3}+...+\frac{1}{81}\right)-\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)}{2}\)
\(\Leftrightarrow X+\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)=\frac{242}{243}:2\)
\(\Leftrightarrow X+\frac{12}{25}=\frac{121}{243}\)
\(\Leftrightarrow X=\frac{109}{6075}\)
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:
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+...+\frac{1}{243}\right)\)
\(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{242}\right)\)
\(12x+\frac{12}{25}=11x+\frac{121}{243}\)
\(12x-11x=\frac{121}{243}-\frac{12}{25}\)
\(x=\frac{109}{6075}\)
\(1\frac{1}{3}\)x \(1\frac{1}{8}\)x \(1\frac{1}{15}\)x .......x \(1\frac{1}{99}\)= \(\frac{4}{3}\)x \(\frac{9}{8}\)x\(\frac{16}{15}\)x..........x \(\frac{100}{99}\)
=\(\frac{2.2}{1.3}\)x\(\frac{3.3}{2.4}\)x\(\frac{4.4}{3.5}\)x.......x \(\frac{10.10}{9.11}\)= \(\frac{2.3.4....10}{1.2.3.....9}\)x \(\frac{2.3.4....10}{3.4.5....11}\)
= \(\frac{10}{1}\)x\(\frac{2}{11}\)=\(\frac{20}{11}\)
4/3 x 9/8 x16/15 x ..... x 100/99
=2x2/1x3 x 3x3/2x4 x 4x4/3x5 x ... x 10x10/9x11
=2x3x4x....x10/1x2x3x..x9 x 2x3x4x...x10/3x4x5x...x11
=10 x 2/11=20/11
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{8}\)
\(=\frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{3.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}+0+0+...+0\)
\(=\frac{7}{8}\)
Bài 2: ta có tích riêng thứ nhất là .....5, thứ hai cũng là ....5 -> chữ số tận cùng là: ....5 - ....5 = ...0
Bài 3: Gọi số có hai chữ số đó là ab (a,b =<9)
...........................__..... _
Theo đề bài ta có: ab = 9b
=> b = (2; 3; 4; 5; 6; 7; 8; 9)
..........................................
=> Tương ứng với b ta có ab = (18; 27; 36; 45; 54; 63; 72; 81)
Nhận xét: Chỉ có 45 = 9.5
Vậy số đó là 45
\(=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x...x\frac{100}{99}\)
\(=\frac{2x2x3x3x4x4x5x5x6x6x7x7x8x8x9x9x10x10}{1x3x2x4x3x5x4x6x5x7x6x8x7x9x8x10x9x11}=\frac{2x10}{11}=\frac{20}{11}\)
Lưu Hạ Băng
Ta có : \(1\frac{1}{3}x1\frac{1}{8}x1\frac{1}{15}x1\frac{1}{24}x...x1\frac{1}{99}\)
\(=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x...x\frac{100}{99}\)
\(=\frac{2x2}{1x3}x\frac{3x3}{2x4}x\frac{4x4}{3x5}x\frac{5x5}{4x6}x...x\frac{10x10}{9x11}\)
\(=\frac{2x2x3x3x4x4x5x5x...x10x10}{1x3x2x4x3x5x4x6x...x9x11}\)
\(=\frac{\left(2x3x4x5x...x10\right)x\left(2x3x4x5x...x10\right)}{\left(1x2x3x4x...x9\right)x\left(3x4x5x6x...x11\right)}\)
\(=\frac{10x2}{1x11}\)
\(=\frac{20}{11}\)
Chúc bn học tốt !!!
Kb vs mk heg =)) ^^