tinh nhanh
\(\left(1-\frac{1}{1x3}\right)x\left(1-\frac{1}{2x4}\right)x\left(1-\frac{1}{3x5}\right)x....x\left(1-\frac{1}{20x22}\right)\)
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ai làm được giúp mình nhé
A= (1-1/1x3)x(1-1/2x4)-(1-1/3x5)......x(1-1/20x22)
\(A=\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right)......\left(1-\frac{1}{1275}\right)\)
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
\(24x+\left(1-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{23}-\frac{1}{25}\right)=22x+\frac{162+54+6+2}{243}\)
\(24x+\frac{24}{25}=22x+\frac{224}{243}\)
\(2x=\frac{224}{243}-\frac{24}{25}\)
\(2x=-\frac{232}{6025}\)
\(x=\frac{-116}{6075}\)
\(\left(x+\frac{1}{1.3}\right)+\left(x+\frac{1}{3.5}\right)+...+\left(x+\frac{1}{23.25}\right)=11.x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{81}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\right)\right]=11.x+\left(\frac{81}{243}+\frac{27}{243}+\frac{3}{243}+\frac{1}{243}\right)\)
\(12x+\left[\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)\right]=11.x+\frac{112}{243}\)
\(12x+\left(\frac{1}{2}.\frac{24}{25}\right)=11.x+\frac{112}{243}\)
\(12x+\frac{12}{25}=11x+\frac{112}{243}\)
\(11x-12x=\frac{112}{243}-\frac{12}{25}\)
\(-1x=-\frac{116}{6075}\)
\(x=-\frac{116}{6075}\div\left(-1\right)\)
\(x=\frac{116}{6075}\)
ta có
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)
\(VT=\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+3}-\frac{1}{x+4}\)
\(=\frac{1}{x}-\frac{1}{x+4}=\frac{x+4-x}{x\left(x+4\right)}=\frac{4}{x\left(x+4\right)}\)
\(\Rightarrow\frac{4}{x\left(x+4\right)}=\frac{m}{x\left(x+4\right)}=VP\Rightarrow m=4\)
quá dễ tách ra thành 1\x-1\x+1+1\x+1-1\x+2+1\x+2-1\x+3+1\x+3-1\x+4+...+1\x+5-1\x+6
=1\x-1\x+6
=6\x(x+6)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\)\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
\(=\frac{1}{x}-\frac{1}{x+6}\)\(=\frac{6}{x\left(x+6\right)}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}=\frac{8}{17}\)
\(\Leftrightarrow2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x\left(x+2\right)}\right)=2.\frac{8}{17}\)
\(\Leftrightarrow\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x\left(x+2\right)}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{16}{17}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{16}{17}=\frac{1}{17}\)
\(\Rightarrow x+2=17\Rightarrow x=15\)