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\(A=\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}+...+\dfrac{1}{87\times89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+...+\dfrac{1}{87}-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-...-\left(\dfrac{1}{87}-\dfrac{1}{87}\right)-\dfrac{1}{89}\)
\(A=\dfrac{1}{5}-\dfrac{1}{89}\)
\(A=\dfrac{84}{445}\)
Vậy, `A=84/445.`
A = \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+\(\dfrac{1}{9\times11}\)+...+\(\dfrac{1}{87\times89}\)
A = \(\dfrac{1}{2}\) \(\times\)( \(\dfrac{2}{5\times7}\)+\(\dfrac{2}{7\times9}\)+\(\dfrac{2}{9\times11}\)+...+\(\dfrac{2}{87\times89}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) + \(\dfrac{1}{11}\) +...+ \(\dfrac{1}{87}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) (\(\dfrac{1}{5}\) - \(\dfrac{1}{89}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{84}{445}\)
A = \(\dfrac{42}{445}\)
A. = 1/2-1/3+1/3-1/4+1/4-1/5...+1/101-1/102=1/2-1/102=25/51.
B. =1/5-1/10+1/10-1/15+...+1/115-1/120=1/5-1/120=23/120.
C. = 1/5-1/7+1/7-1/9+1/9-1/11+...+1/997-1/999=1/5-1/999=994/4995.
Minh kiem tra bang may tinh roi do.
\(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{101\times102}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}\)
\(=1-\frac{1}{102}\)
\(=\frac{101}{102}\)
\(A=11\frac{2}{1.3}+11\frac{2}{3.5}+11\frac{2}{5.7}+11\frac{2}{7.9}+11\frac{2}{9.11}\)
\(A=11+\frac{2}{1.3}+11+\frac{2}{3.5}+11+\frac{2}{5.7}+11+\frac{2}{7.9}+11+\frac{2}{9.11}\)
\(A=55+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(A=55+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(A=55+\frac{1}{1}-\frac{1}{11}\)
\(A=56-\frac{1}{11}\)
\(A=\frac{616}{11}-\frac{1}{11}=\frac{615}{11}\)
Vậy A = 615 / 11
\(\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\left(\frac{1}{1}-\frac{1}{11}\right)\times y=\frac{2}{3}\)
\(\frac{1}{2}\times\frac{10}{11}\times y=\frac{2}{3}\)
\(\frac{5}{11}\times y=\frac{2}{3}\) => \(y=\frac{2}{3}:\frac{5}{11}=\frac{2}{3}\times\frac{11}{5}=\frac{22}{15}\)
\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199