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\(\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}-\dfrac{47}{24}\right)+\dfrac{2}{3}=\dfrac{37}{37}-\dfrac{13}{8}+\dfrac{2}{3}=\dfrac{1}{24}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
\(\dfrac{-5}{9}+1\dfrac{5}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):7^2\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{3}{4}-\dfrac{2}{5}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\left(\dfrac{15}{20}-\dfrac{8}{20}\right):49\\ =\dfrac{-5}{9}+\dfrac{14}{9}.\dfrac{7}{20}:49\\ =\dfrac{-5}{9}+\dfrac{49}{90}:49\\ =\dfrac{-5}{9}+\dfrac{1}{90}\\ =\dfrac{-50}{90}+\dfrac{1}{90}\\ =\dfrac{-49}{90}\)
\(1\dfrac{13}{15}.0,75-\left(\dfrac{104}{195}+25\%\right).\dfrac{24}{47}-3\dfrac{12}{13}:3\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{8}{15}+\dfrac{1}{4}\right).\dfrac{24}{47}-\dfrac{51}{13}:3\\ =\dfrac{7}{5}-\left(\dfrac{32}{60}+\dfrac{15}{60}\right).\dfrac{24}{47}-\dfrac{51}{13}.\dfrac{1}{3}\\ =\dfrac{7}{5}-\dfrac{47}{60}.\dfrac{24}{47}-\dfrac{17}{13}\\ =\dfrac{7}{5}-\dfrac{2}{5}-\dfrac{17}{13}\\ =1-\dfrac{17}{13}\\ =\dfrac{13}{13}-\dfrac{17}{13}\\ =\dfrac{-4}{13}\)
\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)
\(A=\frac{19}{24}+\frac{-1}{2}+\frac{-1}{3}+\frac{-7}{24}\)
\(A=\left(\frac{19}{24}+\frac{-7}{24}\right)+\frac{-1}{2}+\frac{-1}{3}\)
\(A=\frac{1}{2}+\frac{-1}{2}+\frac{-1}{3}\)
\(A=0+\frac{-1}{3}=\frac{-1}{3}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}+\frac{-3}{7}+\frac{-1}{3}\)
\(B=\left(\frac{49}{168}+\frac{140}{168}+\frac{42}{168}\right)+\left(\frac{-72}{168}+\frac{-56}{168}\right)\)
\(B=\frac{231}{168}+\frac{-128}{168}=\frac{103}{168}\)
Có: \(A=\frac{-1}{3}=\frac{\left(-1\right)\cdot56}{3\cdot56}=\frac{-56}{168}\)
Mặt khác: \(-56< 103\)
\(\Rightarrow\)\(\frac{-56}{168}< \frac{103}{168}\)
\(hay\) \(A< B\)
\(-1\dfrac{1}{5}.\dfrac{12+\dfrac{4}{3}-\dfrac{12}{37}-\dfrac{12}{35}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{-6}{5}.\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{35}}:\dfrac{4\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2003}\right)}\)
\(=\dfrac{-6}{5}.4:\dfrac{4}{5}\)
\(=\dfrac{-6.4.5}{5.4}=-6\)
Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )
B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]
B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]
B = \(\dfrac{47}{41}.5\)
B = \(\dfrac{235}{41}\)
Chúc bn hc tốt!!!
mk có thắc mắc là bạn để 3 ra ngoài sao 1/3 vẫn giữ nguyên vậy phải bằng 1/9 mới đúng chứ'
Sửa đề : \(-\dfrac{19}{37}=\dfrac{19}{37}\)
\(\dfrac{19}{37}-\dfrac{31}{24}-\dfrac{19}{37}+1\dfrac{7}{24}-\dfrac{2}{13}\)
\(=\dfrac{19}{37}-\dfrac{31}{24}+\dfrac{19}{37}+\dfrac{31}{24}-\dfrac{2}{13}\)
\(=\left(\dfrac{19}{37}-\dfrac{19}{37}\right)+\left(-\dfrac{31}{24}+\dfrac{31}{24}\right)-\dfrac{2}{13}=-\dfrac{2}{13}\)
ĐỀ ĐÚNG MÀ A