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Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
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= \(\frac{1}{6}--\frac{10}{3}\)[1/6 - (-10/3)]
= \(\frac{7}{2}=3,5\)
\(B=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(B=\frac{2^{12}.3^6-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(B=\frac{2^{12}.3^4.\left(3^2-1\right)}{2^{12}.3^5.\left(3-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(B=\frac{8}{6}-\frac{5-35}{9}\)
\(B=\frac{4}{3}+\frac{30}{9}\)
\(B=\frac{4}{3}+\frac{10}{3}\)
\(B=\frac{14}{3}\)
vậy \(B=\frac{14}{3}\)