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\(\frac{3^3}{6.11}+\frac{3^3}{11.16}+.......+\frac{3^3}{91.96}\)
\(=\frac{3^3}{5}.\left(\frac{5}{6.11}+\frac{5}{11.16}+.....+\frac{5}{91.96}\right)\)
\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+.....+\frac{1}{91}-\frac{1}{96}\right)\)
\(=\frac{27}{5}.\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(=\frac{27}{5}.\frac{5}{32}\)
\(=\frac{27}{32}\)
\(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}\)
\(=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{91}-\frac{1}{96}\right)\)
\(=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)
Vậy \(\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+...+\frac{3^3}{91\cdot96}=\frac{27}{32}\)
a) Đề phải là thế này chứ \(\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{101.106}\)
Giai
\(=\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+...+\frac{5}{101.106}\)
\(=\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{101}-\frac{1}{106}\)
\(=\frac{1}{6}-\frac{1}{106}\)
\(=\frac{25}{159}\)
b) Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5A=1+\frac{1}{5}+...+\frac{1}{5^{99}}\)
\(\Rightarrow5A-A=\left(1+\frac{1}{5}+...+\frac{1}{5^{99}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4A=1-\frac{1}{5^{100}}\)
\(\Rightarrow A=\frac{1-\frac{1}{5^{100}}}{4}\)
\(\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+...+\frac{5^2}{91\cdot96}\)
\(=5\left(\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{5}{91\cdot96}\right)\)
\(=5\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{91}-\frac{1}{96}\right)\)
\(=5\left(\frac{1}{6}-\frac{1}{96}\right)\)\(=5\cdot\frac{5}{32}\)\(=\frac{25}{32}\)
Ta có : (32003 + 32000) : 32000
= 32003 : 32000 + 32000 : 32000
= 33 + 1
= 28
Ta có: \(\left\{3^{2003}+3^{2000}\right\}:3^{2000}\)
\(=3^{2003}:3^{2000}+3^{2000}:3^{2000}\)
\(=3^3+1\)
\(=28\)
Chắc là đúng
S:5=\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{21.26}\)
S:5=\(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{26-21}{21.26}\)
S:5=\(\frac{6}{1.6}-\frac{1}{1.6}+\frac{11}{6.11}-\frac{6}{6.11}+...+\frac{26}{21.26}-\frac{21}{21.26}\)
S:5=1-\(\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{21}-\frac{1}{26}\)
S:5=1-\(\frac{1}{26}\)
S:5=\(\frac{25}{26}\)
S=\(\frac{25}{26}.5\)
S=\(\frac{125}{26}\)
\(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}\)
= \(5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}\right)\)
=\(5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}\right)\)=\(5\left(1-\frac{1}{26}\right)\)
=\(5.\frac{25}{26}\)
=\(\frac{125}{26}\)
\(3^n+3^{n+1}=108\)
\(3^n+3^n.3=108\)
\(3^n.\left(1+3\right)=108\)
\(3^n.4=108\)
\(3^n=108:4\)
\(3^n=27\)
\(3^n=3^3\)
Vậy: \(n=3\)
\( A=\frac{3^3}{6\cdot11}+\frac{3^3}{11\cdot16}+\frac{3^3}{16\cdot21}+....+\frac{3^3}{91\cdot96}\)
\(A=\frac{3^3}{5}\cdot\left(\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+....+\frac{1}{91}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\left(\frac{1}{6}-\frac{1}{96}\right)\)
\(A=\frac{27}{5}\cdot\frac{5}{32}=\frac{27}{32}\)