Tính A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{3×80}\)
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Đáp án: thiếu đề
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mời bn xem xét lại đề bài.
~hok tốt~
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}=\frac{9}{10}\)
Bạn biết bài này rồi phải không =.="
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}\)
\(4+\frac{1}{x}=\frac{4x+1}{x}\)
\(\frac{1}{4+\frac{1}{x}}=\frac{x}{4x+1}\)
\(3+\frac{1}{4+\frac{1}{x}}=3+\frac{x}{4x+1}=\frac{13x+3}{4x+1}\)
Tương tự Vế Trái sẽ tìm đc
\(21+\frac{12\left(13x+3\right)}{30x+7}\)
Vế phải bấm máy tính nhá casio mà
\(VP=\frac{104052}{137}=21+\frac{101175}{137}\)
Suy ra
\(\frac{156x+36}{30x+7}=\frac{101175}{137}\Leftrightarrow21375x+4932=3035250x+708225\)
\(\Leftrightarrow1004625x=-234431\Leftrightarrow x=-\frac{234431}{1004625}\)
A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{3.80}\)
A=\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{240}\)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{15.16}\)
A=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{15}-\frac{1}{16}\)
A=\(1-\frac{1}{16}=\frac{16-1}{16}=\frac{15}{16}\)