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Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)
\(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)
\(=100.\frac{2}{101}=\frac{200}{101}\)
\(A=\frac{2^{19}.\left(2^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{2^9.3^9.2^{10}+\left(2^2.3\right)^{10}}=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}=\frac{2^{18}.3^8.\left(2.3+15\right)}{2^{19}.3^9.\left(1+2.3\right)}\)
\(=\frac{2^{18}.3^8.21}{2^{19}.3^9.7}=\frac{21}{2.3.7}=\frac{1}{2}\)
\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(=\frac{2^{19}.\left(3^3\right)^3+15.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(=\frac{2^{19}.3^9+15.2^{18}.3^8}{2^{19}.3^9+2^{20}.3^{10}}\)
\(=\frac{2^{18}.3^8\left(2.3+15\right)}{2^{19}.3^9\left(1+2.3\right)}\)
\(=\frac{6+15}{2.3\left(1+6\right)}\)
\(=\frac{21}{6.7}\)
\(=\frac{21}{42}\)
\(=\frac{1}{2}\)
Câu 1 :
\(\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^5}\)
= \(\frac{5^{32}.2^{86}.5^{43}}{\left(-2\right)^{87}.5^{15}}\)
= \(\frac{5^{72}.\left(-2\right)^{86}}{\left(-2\right)^{87}.5^{75}}\)
= \(\frac{1}{-2}\)
Câu 2 :
\(\frac{5^4.18^4}{125.9^5.16}\)
= \(\frac{5^4.2^4.3^8}{5^3.3^{10}.2^4}\)
= \(\frac{5}{3^2}\)
= \(\frac{5}{9}\)
Câu 3 :
\(\frac{9^{18}.2^{29}}{8^9.27^{12}}\)
= \(\frac{3^{36}.2^{29}}{2^{27}.3^{36}}\)
= \(2^2\)
= 4
Có P =\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.\left(3^3\right)^3+5.3.\left(3^2\right)^4}{\left(2.3\right)^9+\left(3.2^2\right)^{10}}\)=\(\dfrac{2^{19}.3^9+5.3.2^{18}.3^8}{3^9.2^9.2^{10}+3^{10}.\left(2^2\right)^{10}}=\dfrac{2^{19}.3^9+5.2^{18}.3^9}{3^9.2^{19}+3^{10}.2^{20}}=\dfrac{2^{18}.3^9.\left(2+5\right)}{3^9.2^{19}.\left(1+3.2\right)}=\dfrac{2^{18}.3^9.7}{3^9.2^{19}.7}\)
=\(\dfrac{1}{2}\)
