Question

Giải giúp em mấy bài toán này, gấp nhé. Em cảm ơn

Answer 1

\(1,=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+0,5=1-1+0,5=0,5\\ 2,=-12:\left(-\dfrac{1}{12}\right)^2=12\cdot\dfrac{1}{144}=\dfrac{1}{12}\\ 3,=\dfrac{7}{23}\left(-\dfrac{23}{6}\right)=-\dfrac{7}{6}\\ 4,=\dfrac{7}{5}\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\\ 5,=\dfrac{17}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\\ 6,=\dfrac{5}{3}\left(-16\dfrac{2}{7}+28\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\\ 7,=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=-14\\ 8,=\left[\dfrac{1}{9}\cdot\left(-9\right)\right]^{25}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

\(9,=\dfrac{3}{5}:\left(-\dfrac{7}{30}\right)+\dfrac{3}{5}:\left(-\dfrac{7}{5}\right)=\dfrac{3}{5}\left(-\dfrac{30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\left(-5\right)=-3\\ 10,=5,7\cdot\left(-10\right)=-57\\ 11,=10\cdot\dfrac{1}{10}\cdot\dfrac{4}{3}+21-\dfrac{1}{3}=\dfrac{4}{3}-\dfrac{1}{3}+21=1+21=22\\ 12,=\dfrac{2^{10}}{2^{10}}-\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)