1 Tính
6+25+125+625+....+5200+5201
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\(S=\dfrac{625}{625}+\dfrac{125}{625}+\dfrac{25}{625}+\dfrac{5}{625}+\dfrac{1}{625}\)
\(=\dfrac{781}{625}\)
S = 1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\) + \(\dfrac{1}{625}\)
5.S = 5 +1 + \(\dfrac{1}{5}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{125}\)
5S - S = 5 - \(\dfrac{1}{625}\)
S = ( 5 - \(\dfrac{1}{625}\)) : 4
S = \(\dfrac{781}{625}\)
25%=5201
100%=5201:25*100=20804
(dấu * là dấu nhân nha bạn)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
\(A=1+5+5^2+...+5^{201}\)
\(5A=5+5^2+5^3+...+5^{201}+5^{202}\)
\(4A=5A-A=5^{202}-1\)
\(A=\frac{5^{202}-1}{4}\)
khó,thì sao ? thì vô câu hỏi tương tự thôi