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c) Ta co : A=1/2^1+1/2^2+...+1/2^49+1/2^50
2A=1+1/2+1/2^2+........+1/2^48+1/2^49
A=1-1/2^50<1
Vậy A=1/2^1+1/2^2+...+1/2^49+1/2^50 <1
\(=\left(2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9\right):\left(2^{12}.\left(3^2\right)^3.31\right)\)
\(=\left(2^{13}.3^6+2^{11}.3^9\right):\left(2^{12}.3^6.31\right)\)
\(=\left[2^{11}.3^6\left(2^2+3^3\right)\right]:\left(2^{12}.3^6.31\right)\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
Đưa về phân số:
\(=\frac{2.8^4.27^2+4.6^9}{2^{12}.9^3.31}\)
\(=\frac{2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9}{2^{12}.\left(3^2\right)^3.31}\)
\(=\frac{2.2^{3.4}.3^{3.2}+2^2.2^9.3^9}{2^{12}.3^{2.3}.31}\)
\(=\frac{2.2^{12}.3^6+2^{2+9}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{1+12}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{13}.3^6+2^{11}.3^9}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6\left(2^2+3^3\right)}{2^{12}.3^6.31}\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
Em hiểu hơn ko?
a) \(\dfrac{11\cdot8-11\cdot3}{17-6}\)
\(=\dfrac{11\cdot\left(8-3\right)}{11}=5\)
b) \(\dfrac{24-12\cdot13}{12+4\cdot9}\)
\(=\dfrac{12\cdot\left(2-13\right)}{12\left(1+3\right)}=\dfrac{-11}{4}\)
Lời giải:
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.234.8^2}=\frac{(3^2)^4.(3^3)^5.3^6.3^4}{3^8.(3^4)^4.2.3^2.13.(2^3)^2}\)
\(=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.2.3^2.13.2^6}=\frac{3^{33}}{3^{26}.2^7.13}=\frac{3^7}{2^7.13}\)
\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)
\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)
\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)
\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)
\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)
\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)
\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.243.8^2}\)
\(M=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^6.3^4}{3^8.\left(3^4\right)^4.\left(3^5\right).\left(2^3\right)}\)
\(M=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.3^5.8}\)
\(M=\frac{3^{33}}{3^{29}.8}\)
\(M=\frac{3^4}{1.8}\)
\(M=\frac{81}{8}\)
Chúc bạn học tốt !!!
Bài 1:
\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)
\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)
\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)
a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)