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\(=\left(2.\left(2^3\right)^4.\left(3^3\right)^2+2^2.\left(2.3\right)^9\right):\left(2^{12}.\left(3^2\right)^3.31\right)\)
\(=\left(2^{13}.3^6+2^{11}.3^9\right):\left(2^{12}.3^6.31\right)\)
\(=\left[2^{11}.3^6\left(2^2+3^3\right)\right]:\left(2^{12}.3^6.31\right)\)
\(=\frac{2^{11}.3^6.31}{2^{12}.3^6.31}=\frac{1}{2}\)
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?
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
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.243.8^2}\)
\(M=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^6.3^4}{3^8.\left(3^4\right)^4.\left(3^5\right).\left(2^3\right)}\)
\(M=\frac{3^8.3^{15}.3^6.3^4}{3^8.3^{16}.3^5.8}\)
\(M=\frac{3^{33}}{3^{29}.8}\)
\(M=\frac{3^4}{1.8}\)
\(M=\frac{81}{8}\)
Chúc bạn học tốt !!!
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
A,
\(\left(7\dfrac{4}{9}+3\dfrac{7}{11}\right)-3\dfrac{4}{9}=7\dfrac{4}{9}+3\dfrac{7}{11}-3\dfrac{4}{9}\)
\(=7\dfrac{4}{9}-3\dfrac{4}{9}+3\dfrac{7}{11}=4+3\dfrac{7}{11}=7\dfrac{7}{11}\)
B,
\(5\dfrac{2}{7}.\dfrac{8}{11}+5\dfrac{2}{7}.\dfrac{5}{11}-5\dfrac{2}{7}.\dfrac{2}{11}=5\dfrac{2}{7}.\left(\dfrac{8}{11}+\dfrac{5}{11}-\dfrac{2}{11}\right)\)
\(=5\dfrac{2}{7}.1=5\dfrac{2}{7}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(\frac{16.3^{11+4.27^4}}{9^7}=\frac{2^4.3^{11+2^2.3^{12}}}{3^{14}}\)
\(=\frac{2^4.3^{11+2^2.531441}}{3^{14}}=\frac{2^4.3^{2125764}}{3^{14}}=2^4.3^{2125750}\)
Ta có:
\(\frac{16.3^{11+4.27^4}}{9^7}=\frac{2^4.3^{11+2^2.\left(3^3\right)^4}}{\left(3^2\right)^7}=\frac{2^4.3^{11+2^2.3^{12}}}{3^{14}}=\frac{2^4.3^{11}.3^{2^2.3^{12}}}{3^{14}}=\frac{2^4.3^{15}.3^{3^{12}}}{3^{14}}=2^4.3^{3^{12}}\)
Vậy ...