TÌM GIÁ TRỊ BIỂU THỨC A=1/4+1/12+1/24+1/40+1/60+1/84.HELL
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\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)NHÂN CẢ TỬ VÀ MẪU CỦA TỪNG P/S VỚI 2 TA ĐƯỢC:
\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)
\(A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{12.14}\)
\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{12}-\frac{1}{14}\)
\(A=\frac{1}{2}-\frac{1}{14}\)
\(A=\frac{3}{7}\)
A = 1/4 + ... +1/84
A = 2/8 + 2/24 + ... + 2/168
A = 2/2.4 + 2/4.6 + ... + 2/12.14
A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14
A = 1/2 - 1/14
A = 6/14 = 3/7
A = 2/8 + 2/24 + ... + 2/168
A = 2/2.4 + 2/4.6 + ... + 2/12.14
A = 1/2 - 1/4 + 1/4 - 1/6 + .. + 1/12 - 1/14
A = 1/2 - 1/14
A = 3/7
2A=1/2+1/6+1/12+1/20+1/30+1/42
=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1-1/7
=6/7
\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}+x=1\)
\(\)\(2\cdot\left(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\right):2+x=1\)
\(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right):2+x=1\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right):2+x=1\)
\(\dfrac{6}{7}:2+x=1\)
\(x=1-\dfrac{3}{7}\)
\(x=\dfrac{4}{7}\)
\(\)
\(\dfrac{1}{4}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{40}\) + \(\dfrac{1}{60}\) + \(\dfrac{1}{84}\) + \(x\) = 1
\(\dfrac{1}{2.2}\) + \(\dfrac{1}{2.6}\)+\(\dfrac{1}{2.12}\)+\(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.30}\) + \(\dfrac{1}{2.42}\) + \(x\) =1
\(\dfrac{1}{2}\).(\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{42}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).( \(\dfrac{1}{1}\) - \(\dfrac{1}{7}\)) + \(x\) = 1
\(\dfrac{1}{2}\).\(\dfrac{6}{7}\) + \(x\) = 1
\(\dfrac{3}{7}\) + \(x\) = 1
\(x\) = 1 - \(\dfrac{3}{7}\)
\(x\) = \(\dfrac{4}{7}\)
\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\\ =\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{80}+\dfrac{2}{120}+\dfrac{2}{168}\\ =\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{6}{14}\\ =\dfrac{3}{7}\)
\(\dfrac{1}{4}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{84}\\ =\dfrac{2}{8}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{80}+\dfrac{2}{120}+\dfrac{2}{168}\\ =\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}+\dfrac{2}{10\cdot12}+\dfrac{2}{12\cdot14}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}\\ =\\ 1-\dfrac{1}{14}\\ =\dfrac{14}{14}-\dfrac{1}{14}\\ =\dfrac{13}{14}\)