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4 tháng 12 2023

\(A=\dfrac{6+3^2.3-6+3^2}{3^2+3.3^2-3^4}=\dfrac{\left(6-6\right)+3^2\left(3+1\right)}{3^2\left(1+3-3^2\right)}\\ =\dfrac{0+3^2.4}{3^2.\left(-5\right)}=\dfrac{-4}{5}\)

18 tháng 7 2019

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

18 tháng 7 2019

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

\(=\frac{3^4}{3^3}\)

\(=3\)

Sửa đề: \(C=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\right)^6\cdot3^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(C=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}\)

\(=\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-6\right)}{9}\)

\(=\dfrac{2}{12}+\dfrac{30}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{1}{6}+\dfrac{20}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)

13 tháng 12 2017

a) \(2010^{100}+\)\(2010^{99}=2010^{99}.2010+2010^{99}.1=2010^{99}.\left(2010+1\right)=2010^{99}.2011\)Vậy biểu thức chia hết cho 2011.

13 tháng 1 2018

\(\dfrac{2^{12}.3^5-4^6.3^6}{2^{12}.9^3+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^6.3}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^5}.\)

\(=\dfrac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}\left(3^5-3^6\right)}{2^{12}\left(3^5+3^6\right)}=\dfrac{3^5-3^6}{3^5+3^6}=-\dfrac{486}{972}=-\dfrac{1}{2}.\)

Vậy..........

23 tháng 12 2018

\(a)\dfrac{1}{4}-\dfrac{3}{4}:\left(\dfrac{-5}{8}\right)\)

\(=\dfrac{1}{4}-\dfrac{3}{4}.\dfrac{-8}{5}\)

\(=\dfrac{1}{4}-\dfrac{-6}{5}\)

\(=\dfrac{5}{20}+\dfrac{24}{20}\)

\(=\dfrac{29}{20}\)

\(b)3-\left(\dfrac{-6}{7}\right)^0+\sqrt{\dfrac{1}{16}}:2\)

\(=3-1+\sqrt{\left(\dfrac{1}{4}\right)^2}:2\)

\(=2+\dfrac{1}{4}.\dfrac{1}{2}\)

\(=\dfrac{16}{8}+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

\(c)\dfrac{9^5.2^6}{4^3.3^8}=\dfrac{\left(3^2\right)^5.2^6}{\left(2^2\right)^3.3^8}=\dfrac{3^{10}.2^6}{2^6.3^8}=3^2=9\)

23 tháng 12 2018

cảm ơn bạn

16 tháng 7 2018

\(a)A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^63.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)

\(A=\dfrac{2^{12}.3^5-2^{12}.3^5}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.7^3.2^3}\)

\(A=\dfrac{0}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+5^3+2^3\right)}\)

\(A=0-\dfrac{5^4.\left(-6\right)}{1+125+8}\)

\(A=0-\dfrac{625.\left(-6\right)}{134}\)

\(A=\dfrac{-3750}{134}\)\(=\dfrac{-1875}{67}\)

\(b)3^{n+2}-2^{n+2}+3^n-2^n\)

\(=3^n.3^2-2^n.2^2+3^n-2^n\)

\(=(3^n.9+3^n)-\left(2^n.4+2^n\right)\)

\(=3^n.10-2^n.5\)

\(=3^n.10-2^{n-1}.10\)

\(=10\left(3^n-2^{n-1}\right)⋮10\)

\(Suy\) \(ra:\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)

16 tháng 7 2018

b. Ta có: \(3^{n +2}-2^{n+2}+3^n-2^n\)

\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

\(=\left(3^n.3^2+3^n\right)-\left(2^{n-1}.2^3+2^{n-1}.2\right)\)

\(=3^n.\left(3^2+1\right)-2^{n-1}\left(2^3+2\right)\)

\(=3^n.10-2^{n-1}.10⋮10\)

20 tháng 11 2016

\(\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^3}\)

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

\(\frac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^3.8}\)

\(\frac{3^2.\left(-1\right)}{4}\)

\(\frac{-9}{4}\)

VẬy.......................

nhớ tk cho mình nha