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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)
=\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}\left(3^6+3^5\right)}-\frac{5^{10}\left(7^3-7^4\right)}{5^9.7^3\left(1+2^3\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}-\frac{5\left(7^3-7^4\right)}{7^3.3^2}\)
=\(\frac{3^4\left(3-1\right)}{^{ }3^4\left(9+3\right)}-\frac{5.7^3-5.7^4}{7^3.3^2}\)
=\(\frac{1}{6}-\frac{7^3.5\left(1-7\right)}{7^3.3^2}=\frac{1}{6}-\frac{30}{9}=-\frac{19}{6}\)
Vậy A=\(-\frac{19}{6}\)
câu b lúc nã mk làm sai rui
dây mới đúng
=\(\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{96}-\frac{1}{101}\right)\)
=\(\frac{1}{5}\left(1-\frac{1}{101}\right)=\frac{1}{5}.\frac{100}{101}=\frac{20}{101}\)
\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot8}\)
\(=\dfrac{3^4\cdot\left(3-1\right)}{3^5\left(3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}=\dfrac{2}{3\cdot4}-\dfrac{5\cdot\left(-2\right)}{3}\)
\(=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{1}{6}+\dfrac{20}{6}=\dfrac{21}{6}=\dfrac{7}{2}\)
mk làm bài 1 thui,bài 2 chỉ qui đồng ms
3a/6 = 3b/4 => 3(a-b)/ (6-4) = 3.4,5/2= 13,5/2 =k
a = 2k=13,5
b = 4k/3 =9