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\(A=\frac{5}{13}+\frac{-5}{7}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(\Leftrightarrow A=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=1+\left(-1\right)+\frac{-5}{7}\)
\(\Leftrightarrow A=0+\frac{-5}{7}=\frac{-5}{7}\)
Vậy A = \(\frac{-5}{7}\)
B= \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(\Leftrightarrow B=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\Leftrightarrow B=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(\Leftrightarrow B=-1+1+\frac{-2}{11}\)
\(\Leftrightarrow B=0+\frac{-2}{11}\)
\(\Leftrightarrow\) \(B=\frac{-2}{11}\)
Vậy \(B=\frac{-2}{11}\)
@@ Học tốt
Chiyuki Fujito
K cần tk nhá
\(\frac{17}{23}.\frac{8}{16}.\frac{23}{17}.\left(-80\right).\frac{3}{4}\)
\(=\left(\frac{17}{23}.\frac{23}{17}\right).\left(\frac{1}{2}.\frac{3}{4}.\left(-80\right)\right)\)
\(=1.\left(\frac{3}{8}.\left(-80\right)\right)\)
\(=1.\left(-30\right)=-30\)
\(A=\left(2^2+2^3+2^4+2^5 \right).\left(3^2+3^3+3^4\right)\left(2^4-4^2\right)\)
\(=\left(2^2+2^3+2^4+2^5\right).\left(3^2+3^3+3^4\right).\left(16-16\right)\)
\(=0\)
\(18.26.25.9=2.2.25.9.9.13=100.81.13=105300\)
\(24.3.5.10=3.3.8.5.10=9.400=3600\)
\(14.25.6.7=7.2.25.2.3.7=49.3.100=14700\)