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Rút gọn biểu thức sau:
\(B=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-3\)
\(\Rightarrow B=\dfrac{3^{101}-3}{2}\)
3^x=3^(1+2+3+....+100)
x=1+2+3+..+100
x=(100+1).100/2=5050
\(3^x=3^1\cdot3^2\cdot...\cdot3^{100}\)
\(3^x=3^{1+2+...+100}\)
\(3^x=3^{5050}\)
Vậy x = 5050
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}\)
\(A=\left(\frac{2.3....99}{1.2....98}\right).\left(\frac{2.3....99}{3.4....100}\right)\)
\(A=\frac{99}{1}.\frac{2}{100}\)
\(A=\frac{198}{100}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)
\(=\frac{5.4-2}{19}=\frac{18}{19}\)
Bài giải
a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)
\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)
b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)
B=31+2+..+100=35050
đấy là dấu nhân hay cộng