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a)\(\frac{3^{10}.\left(-5\right)^{21}}{\left(-5\right)^{20}.3^{12}}=\frac{-5}{9}\)
b)\(\frac{\left(-11\right)^5.13^7}{11^5.13^8}=-\frac{1}{13}\)
c)\(\frac{2^{10}.3^{10}-2^{10}.3^9}{2^9.3^{10}}=\frac{2^{10}.3^9\left(3-1\right)}{2^9.3^{10}}=2\)
d(\(\frac{5^{11}.7^{12}+5^{11}.7^{11}}{5^{12}.7^{12}+9.5^{11}.7^{11}}=\frac{5^{11}.7^{11}\left(7+1\right)}{5^{11}.7^{11}\left(35+9\right)}=\frac{1}{6}\)
\(A=\dfrac{10^{12}+6}{10^{12}-11}\)
\(\Rightarrow A=\dfrac{10^{12}-11+17}{10^{12}-11}\)
\(\Rightarrow A=\dfrac{10^{12}-11}{10^{12}-11}+\dfrac{17}{10^{12}-11}\)
\(\Rightarrow A=1-\dfrac{17}{10^{12}-11}\)
\(B=\dfrac{10^{11}+5}{10^{11}-12}\)
\(\Rightarrow B=\dfrac{10^{11}-12+17}{10^{11}-12}\)
\(\Rightarrow B=\dfrac{10^{11}-12}{10^{11}-12}+\dfrac{17}{10^{11}-12}\)
\(\Rightarrow B=1-\dfrac{17}{10^{11}-12}\)
Vậy ta cần so sánh \(1-\dfrac{17}{10^{12}-11}\) và \(1-\dfrac{17}{10^{11}-12}\)
Ta thấy \(\left(10^{12}-11\right)>\left(10^{11}-12\right)\) và 2 phân số trên cùng tử số 17 nên \(\dfrac{17}{10^{12}-11}< \dfrac{17}{10^{11}-12}\)
Vậy \(1-\dfrac{17}{10^{12}-11}>1-\dfrac{17}{10^{11}-12}\) hay \(A>B\)
\(a,-\frac{4}{7}+\left(-\frac{2}{3}+\frac{3}{7}\right)=-\frac{4}{7}-\frac{2}{3}+\frac{3}{7}=-\frac{4}{7}+\frac{3}{7}-\frac{2}{3}=-\frac{1}{7}-\frac{2}{3}=-\frac{17}{21}\)
\(b,-\frac{3}{10}\cdot\frac{5}{12}+-\frac{3}{10}\cdot\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot-\frac{5}{12}+\frac{3}{10}\cdot-\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot\left(-\frac{5}{12}-\frac{7}{12}+5\right)\)
\(=\frac{3}{10}\cdot4=\frac{6}{5}\)
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\(\frac{-4}{7}+\left(\frac{-2}{3}+\frac{4}{7}\right)\)
\(=\frac{-4}{7}+\frac{-2}{3}+\frac{4}{7}\)
\(=\left(\frac{-4}{7}+\frac{4}{7}\right)+\frac{-2}{3}\)
\(=0+\frac{-2}{3}=\frac{-2}{3}\)
~ Hok tốt ~
\(\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+5\frac{3}{10}\)
\(=\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+\frac{53}{10}\)
\(=\frac{-3}{10}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{53}{10}\)
\(=\frac{-3}{10}.1+\frac{53}{10}\)
\(=\frac{-3}{10}+\frac{53}{10}=\frac{50}{10}=5\)
~ Hok tốt ~