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1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=-12:\left(\dfrac{-1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)
\(=-1728\)
c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)
\(=\dfrac{7}{23}.\dfrac{-23}{6}\)
\(=\dfrac{-7}{6}\)
d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)
\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)
\(=10.\dfrac{7}{5}\)
\(=14\)
e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)
\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)
\(=\left(-\dfrac{3}{4}+\dfrac{5}{13}\right).\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right).\dfrac{7}{2}\\ =\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{12}{4}-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{39}{13}-\dfrac{3}{13}\right)\\ =\dfrac{7}{2}\left(-\dfrac{42}{13}\right)\\ =\dfrac{7.\left(-42\right)}{2.13}\\ =-\dfrac{147}{13}\)
\(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}:\dfrac{5}{7}\)
= \(\dfrac{93}{4}.\dfrac{7}{5}-\dfrac{53}{4}.\dfrac{7}{5}\)
= \(\dfrac{7}{5}.\left(\dfrac{93}{4}-\dfrac{53}{4}\right)\)
= \(\dfrac{7}{5}.10\)
= \(14\)
( 2x -1)2 - \(\dfrac{1}{4}\) = 2 ( -2\(\dfrac{1}{5}\) + \(\dfrac{2}{7}\) - \(\dfrac{12}{13}\)) - ( - \(\dfrac{5}{7}\)+\(\dfrac{1}{13}\))
(2x -1)2 = 2 + \(\dfrac{1}{4}\) = -\(\dfrac{11}{5}\) + \(\dfrac{2}{7}\) - \(\dfrac{12}{13}\) + \(\dfrac{5}{7}\) - \(\dfrac{1}{13}\)
(2x -1) 2 = \(\dfrac{9}{4}\) = - \(\dfrac{11}{5}\) + ( \(\dfrac{2}{7}+\dfrac{5}{7}\)) - ( \(\dfrac{12}{13}+\dfrac{1}{13}\))
\(\left[{}\begin{matrix}2x-1=\dfrac{3}{2}\\2x-1=-\dfrac{3}{2}\end{matrix}\right.\) = - \(\dfrac{11}{5}\) + 1 - 1
\(\left[{}\begin{matrix}2x=\dfrac{3}{2}+1\\2x=-\dfrac{3}{2}+1\end{matrix}\right.\) = -11/5
\(\left[{}\begin{matrix}2x=2,5\\2x=-0,5\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1,25\\x=-0,25\end{matrix}\right.\)
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{264}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
Sửa đề
\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)
A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)
=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)
Sửa đề:
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\)
\(A=\dfrac{3}{5}.\dfrac{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}{\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{2}.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}\right)}\)
\(A=\dfrac{3}{5}+\dfrac{1}{\dfrac{5}{2}}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
Chúc bạn hcọ tốt!!!
\(A=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{\dfrac{3}{7}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{2}{4}+\dfrac{2}{8}-\dfrac{2}{6}}{\dfrac{5}{4}+\dfrac{5}{8}-\dfrac{5}{6}}=\dfrac{3\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}{5\left(\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{13}\right)}+\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{6}\right)}{5\left(\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{7}\right)}=\dfrac{3}{5}+\dfrac{2}{5}=1\)
ko dùng mt thì lâu ls đó, phải giải từ từ ra.
Theo đề của bạn ta được: biến thành phép chia phân số:
\(\left(\dfrac{4}{13}-0,8+\dfrac{4}{7}+1\right):\left(\dfrac{13}{7}-2,6+1+3,25\right)\)
\(=\left(\dfrac{4}{13}-0,8+\dfrac{4}{7}+1\right).\left(\dfrac{7}{13}-\dfrac{10}{26}+1+\dfrac{100}{325}\right)\)
Quy đồng rồi nhân ta được:
\(\dfrac{4}{13}\)
cái này bấm máy tính là ra:
\(\dfrac{4}{13}\)