B=3^1+2+..+100=3⁵⁰⁵⁰

đấy là dấu nhân hay cộng

\(3B=3^2.3^3.....3^{101}\)

\(3B-B=3^{101}-3\)

\(2B=3^{101}-3\)

\(B=\frac{3^{101}-3}{2}\)

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

\(\text{Tìm }x\text{ biết : }\)

\(3^x=3\cdot3^2\cdot3^3\cdot3^4\cdot3^5\cdot...\cdot3^{99}\cdot3^{100}\)

\(3^x=3^1\cdot3^2\cdot3^3\cdot3^4\cdot3^5\cdot...\cdot3^{99}\cdot3^{100}\)

\(3^x=3^{1+2+3+4+5+...+99+100}\)

\(3^x=3^{\left(100+1\right)\cdot\left[\left(100-1\right)\text{ : }1+1\right]\text{ : }2}\)

\(3^x=3^{101\cdot\left(99\text{ }\text{ : }1+1\right)\text{ : }2}\)

\(3^x=3^{101\cdot100\text{ : }2}\)

\(3^x=3^{10100\text{ : }2}\)

\(3^x=3^{5050}\)

\(\Rightarrow\text{ }x=5050\)

\(Q=\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}=\frac{2}{3.4}=\frac{1}{6}\)

Q = \(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)

    = \(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

    = \(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}\)

    = \(\frac{2}{3.4}=\frac{1}{6}\)

\(Q=\frac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\)

\(Q=\frac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{2^{12}\cdot3^6+\left(2^3\right)^4\cdot3^5}\)

\(Q=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}\)

\(Q=\frac{2^{12}\cdot\left(3^5-3^4\right)}{2^{12}\cdot\left(3^6+3^5\right)}\)

\(Q=\frac{3^5-3^4}{3^6+3^5}\)

\(Q=\frac{162}{972}\)

\(Q=\frac{81}{486}\)

\(Q=\frac{1}{6}\)

\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}....\frac{99.99}{98.100}\)

\(A=\left(\frac{2.3....99}{1.2....98}\right).\left(\frac{2.3....99}{3.4....100}\right)\)

\(A=\frac{99}{1}.\frac{2}{100}\)

\(A=\frac{198}{100}\)

e chịu thui

\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)

\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)

\(A=1.2.3.4...2019.\left(2020.2021-2020^2\right)=1.2.3.4...2019.2020\)

\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)

\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)

\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)

\(=\frac{5.4-2}{19}=\frac{18}{19}\)

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