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a: \(\left(\dfrac{5}{6}\right)^6\cdot\left(\dfrac{6}{5}\right)^6\cdot\left(\dfrac{6}{5}\right)^2=\left(\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^6\cdot\dfrac{36}{25}=\dfrac{36}{25}\)
b: \(=-\left(\dfrac{13}{8}\right)^3\cdot\left(\dfrac{32}{13}\right)^3\cdot\dfrac{32}{13}\)
\(=-\left(\dfrac{13}{8}\cdot\dfrac{32}{13}\right)^3\cdot\dfrac{32}{13}=-4^3\cdot\dfrac{32}{13}=\dfrac{-2048}{13}\)
c: \(=\left(0.1\right)^7\cdot10^{13}=\left(0.1\cdot10\right)^7\cdot10^6=10^6\)
Giải:
1) \(7^8.\left(-\dfrac{1}{7}\right)^8\)
\(=7^8.\left(\dfrac{1}{7}\right)^8\)
\(=7^8.\dfrac{1^8}{7^8}\)
\(=1\)
2) \(\left(\dfrac{4}{3}\right)^{10}.\left(-\dfrac{3}{4}\right)^{10}\)
\(=\left(\dfrac{4}{3}\right)^{10}.\left(\dfrac{3}{4}\right)^{10}\)
\(=\dfrac{4^{10}}{3^{10}}.\dfrac{3^{10}}{4^{10}}\)
\(=1\)
3) \(\left(-\dfrac{7}{2}\right)^{2006}.\left(-\dfrac{2}{7}\right)^{2006}\)
\(=\left(\dfrac{7}{2}\right)^{2006}.\left(\dfrac{2}{7}\right)^{2006}\)
\(=1\)
4) \(\left(-\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\left(\dfrac{5}{13}\right)^{2007}.\left(\dfrac{13}{5}\right)^{2006}\)
\(=\dfrac{5^{2007}.13^{2006}}{13^{2007}.5^{2006}}\)
\(=\dfrac{5}{13}\)
Vậy ...
A = \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{4}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(=\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(=\frac{3}{5}+\frac{2}{5}=1\)
b) B = \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6.8^4.3^5}-\frac{5^{10}.7^3:25^5.49}{\left(125.7\right)^3+5^9.14^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.7^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}-7^2}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^2.\left(7-1\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{1}{3.2}-\frac{5.2}{7.3}\)
\(=\frac{7}{3.2.7}-\frac{5.2.2}{7.3.2}\)
\(=\frac{7}{42}-\frac{20}{42}\)
\(=-\frac{13}{42}\)
a, 6/7 + (2/11 - 6/7) - (13/11 + 1)
= 6/7 + 2/11 - 6/7 - 13/11 - 1
= (6/7 - 6/7) - (13/11 - 2/11) - 1
= 0 - 1 - 1
= -2
a, \(\frac{\left(\frac{1}{9}\right)^6\cdot\left(\frac{3}{8}\right)^7}{\left(\frac{1}{3}\right)^{13}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{\left(\frac{1}{\left(3^2\right)^6}\right)\cdot\left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot3\right)^7}{\left(\frac{1}{3}\right)^{13}.\left(\frac{1}{2}\right)^{22}.3^6}=\frac{\frac{1}{3^{12}}\cdot\left(\frac{1}{2}\right)^{21}\cdot3^7}{\frac{1}{3^{13}}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{3}{\frac{1}{3}\cdot\frac{1}{2}}=3\div\frac{1}{6}=3.6=18\)
b, Làm tương tự nha bn
a= -1
b= (4)^12
c=1
banj ơi viết rõ ra giùm mình với