Rút gọn biểu thức sau:
B = \(3\cdot3^2\cdot3^3\cdot3^4\cdot...\cdot3^{100}\).
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\(B=3+3^2+3^3+...+3^{100}\)
\(\Rightarrow3B=3^2+3^3+3^4+...+3^{101}\)
\(\Rightarrow3B-B=\left(3^2+3^3+3^4+...+3^{101}\right)-\left(3+3^2+3^3+...+3^{100}\right)\)
\(\Rightarrow2B=3^{101}-3\)
\(\Rightarrow B=\dfrac{3^{101}-3}{2}\)
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
\(Q=\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}=\frac{2}{3.4}=\frac{1}{6}\)
Q = \(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
= \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
= \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}\)
= \(\frac{2}{3.4}=\frac{1}{6}\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}\)
\(A=\left(\frac{2.3....99}{1.2....98}\right).\left(\frac{2.3....99}{3.4....100}\right)\)
\(A=\frac{99}{1}.\frac{2}{100}\)
\(A=\frac{198}{100}\)
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)
\(=\frac{5.4-2}{19}=\frac{18}{19}\)
\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)
\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)
B=31+2+..+100=35050
đấy là dấu nhân hay cộng