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1: \(=\dfrac{15}{37}\cdot\dfrac{38}{41}-\dfrac{15}{37}\cdot\dfrac{74}{45}-\dfrac{38}{41}\cdot\dfrac{15}{37}-\dfrac{38}{41}\cdot\dfrac{82}{76}\)
\(=\dfrac{-2}{3}-1=-\dfrac{5}{3}\)
2: \(=\dfrac{47}{53}\cdot\dfrac{17}{3}-\dfrac{47}{53}\cdot\dfrac{53}{47}+\dfrac{17}{3}\cdot\dfrac{6}{17}-\dfrac{17}{3}\cdot\dfrac{47}{53}\)
\(=-1+2=1\)
9: \(=\dfrac{47}{51}\cdot\dfrac{17}{94}-\dfrac{47}{51}\cdot\dfrac{53}{91}-\dfrac{53}{91}\cdot\dfrac{91}{53}+\dfrac{53}{91}\cdot\dfrac{47}{51}\)
\(=\dfrac{1}{6}-1=-\dfrac{5}{6}\)
10: \(=\dfrac{13}{19}\cdot\dfrac{19}{26}-\dfrac{13}{19}\cdot\dfrac{71}{43}+\dfrac{71}{43}\cdot\dfrac{13}{19}-\dfrac{71}{43}\cdot\dfrac{86}{71}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)
b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)
c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)
\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
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'
\(=\frac{47}{53}.\frac{17}{3}-\frac{47}{53}.\frac{53}{47}+\frac{17}{3}.\frac{6}{17}-\frac{17}{3}.\frac{47}{53}\)
\(=\left(\frac{47}{53}.\frac{17}{3}-\frac{17}{3}.\frac{47}{53}\right)+\left(\frac{47}{53}.\frac{53}{47}-\frac{17}{3}.\frac{6}{17}\right)\)
\(=0+\left(-1\right)\)
\(=-1\).
Giải:
a) \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)
\(=\dfrac{1}{100}\)
Vậy giá trị của biểu thức trên là \(\dfrac{1}{100}\).
b) \(\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{4}{11}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\left(\dfrac{44}{121}+\dfrac{3}{121}-\dfrac{47}{121}\right)\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).\dfrac{0}{121}\)
\(=\left(\dfrac{2}{175}-\dfrac{7}{25}+\dfrac{3}{5}\right).0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
c) \(-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}\right)-\dfrac{2}{5}\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left[\left(\dfrac{15}{17}-\dfrac{9}{15}\right)+\left(\dfrac{2}{17}+\dfrac{-2}{5}\right)\right]\)
\(=-\dfrac{2}{5}\left(\dfrac{15}{17}-\dfrac{9}{15}+\dfrac{2}{17}+\dfrac{-2}{5}\right)\)
\(=-\dfrac{2}{5}\left(1-1\right)\)
\(=-\dfrac{2}{5}.0\)
\(=0\)
Vậy giá trị của biểu thức trên là 0.
Chúc bạn học tốt!!!
\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^3}{3.4}...\dfrac{99^2}{99.100}\)
\(=\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}....\dfrac{99.99}{99.100}\)
\(=\dfrac{1.1.2.2.3.3.....99.99}{1.2.2.3.3.4....99.100}\)
\(=\dfrac{1.2.3...99}{1.2.3....99}.\dfrac{1.2.3....99}{2.3.4....100}=1.\dfrac{1}{100}=\dfrac{1}{100}\)
\(\dfrac{47}{53}\left(\dfrac{17}{3}-\dfrac{53}{47}\right)+\dfrac{17}{3}\left(\dfrac{6}{17}-\dfrac{47}{53}\right)\)
\(=\dfrac{47}{53}.\dfrac{17}{3}-\dfrac{45}{53}.\dfrac{53}{47}+\dfrac{17}{3}.\dfrac{6}{17}-\dfrac{17}{3}.\dfrac{47}{53}\)
\(=-1+\dfrac{6}{3}=-1+2=1\)
Hình như bạn viết sai đề, phân số đầu tiên có thể là 47/53