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Tình hợp lí
C=(11213.20 +11220.27 +11227.34 +...+11262.69 ):(−59.13 −79.25 −1319.25 −3119.69 )
c) Ta co : A=1/2^1+1/2^2+...+1/2^49+1/2^50
2A=1+1/2+1/2^2+........+1/2^48+1/2^49
A=1-1/2^50<1
Vậy A=1/2^1+1/2^2+...+1/2^49+1/2^50 <1
\(M=\dfrac{9}{40}-\dfrac{11}{60}+\dfrac{13}{84}-\dfrac{15}{112}\)
\(M=\left(\dfrac{90}{40}-\dfrac{11}{60}\right)+\left(\dfrac{13}{84}-\dfrac{15}{112}\right)\)
\(M=\dfrac{27}{120}-\dfrac{22}{120}+\dfrac{52}{336}-\dfrac{45}{336}\)
\(M=\dfrac{5}{120}+\dfrac{7}{336}\)
\(M=\dfrac{1}{24}+\dfrac{1}{48}\)
\(M=\dfrac{3}{48}=\dfrac{1}{16}\)
M=\(\dfrac{9}{40}-\dfrac{11}{60}+\dfrac{13}{84}-\dfrac{15}{112}\)
M=\(\dfrac{1}{24}+\dfrac{13}{84}-\dfrac{15}{112}\)
M=\(\dfrac{11}{56}-\dfrac{15}{112}\)
M=\(\dfrac{1}{16}\)
Ta có: \(M=\dfrac{9}{40}-\dfrac{11}{60}+\dfrac{13}{84}-\dfrac{15}{112}\)
\(=\dfrac{378}{1680}-\dfrac{308}{1680}+\dfrac{260}{1680}-\dfrac{225}{1680}\)
\(=\dfrac{105}{1680}=\dfrac{1}{16}\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
\(\dfrac{3}{7}.\left(\dfrac{13}{8}-\dfrac{7}{9}\right)-\dfrac{5}{8}.\left(\dfrac{3}{7}-\dfrac{8}{15}\right)\)
\(=\dfrac{3}{7}.\dfrac{13}{8}-\dfrac{3}{7}.\dfrac{7}{9}-\dfrac{5}{8}.\dfrac{3}{7}-\dfrac{5}{8}.\dfrac{8}{15}\)
\(=\dfrac{3}{7}.\left(\dfrac{13}{8}-\dfrac{5}{8}\right)-\dfrac{3.7}{7.9}-\dfrac{5.8}{8.15}\)
\(=\dfrac{3}{7}.\dfrac{8}{8}-\dfrac{1.1}{1.3}-\dfrac{1.1}{1.3}\)
\(=\dfrac{3}{7}.1-\dfrac{1}{3}-\dfrac{1}{3}\)
\(=\dfrac{3}{7}-\dfrac{1}{3}-\dfrac{1}{3}\)
\(=\dfrac{3}{7}-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}-0\)
\(=\dfrac{3}{7}\)