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a) A = 15/12 + 5/13 + (-3/12) + (-18/13)
= (15/12 - 3/12) + (5/13 - 18/13)
= 1 - 1
= 0
b) B = 11/15 . (-19/13) + (-7/13) . 11/15
= 11/15.(-19/13 - 7/13)
= 11/15 . (-2)
= -22/15
c) C = 2022⁰ - (1/7)⁵ . 7⁵
= 1 - 1/7⁵ . 7⁵
= 1 - 1
= 0
\(A=\dfrac{\left(1+17\right).\left(1+\dfrac{17}{2}\right)..........\left(1+\dfrac{17}{19}\right)}{\left(1+19\right).\left(1+\dfrac{19}{2}\right)..........\left(1+\dfrac{19}{17}\right)}\)
\(=\dfrac{18.\dfrac{19}{2}.............\dfrac{36}{19}}{20.\dfrac{21}{2}..........\dfrac{36}{17}}\)
\(=\dfrac{18.19.20.......36}{1.2.3...19}:\dfrac{20.21.....36}{1.2.3...17}\)
\(=\dfrac{1.2.3......36}{1.2.....36}\)
\(=1\)
a: \(=\dfrac{3}{8}\left(27+\dfrac{1}{5}-51-\dfrac{1}{5}\right)+19\)
\(=-24\cdot\dfrac{3}{8}+19=-9+19=10\)
b: \(=\left(35+\dfrac{1}{6}-46-\dfrac{1}{6}\right):\left(\dfrac{-4}{5}\right)\)
\(=\dfrac{-11\cdot5}{-4}=\dfrac{55}{4}\)
c: \(=\left(\dfrac{-15+8}{20}\right):\left[\dfrac{3}{7}+\dfrac{7}{3}\cdot\dfrac{12-5}{20}\right]\)
\(=\dfrac{-7}{20}:\left(\dfrac{3}{7}+\dfrac{49}{60}\right)\)
\(=-\dfrac{147}{523}\)
Ta có: \(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\dfrac{8}{5}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{32}{65}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}=\dfrac{10}{13}\cdot\dfrac{65}{32}=\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}-1=\dfrac{1}{4}\\2x=-\dfrac{5}{4}-1=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:2=\dfrac{1}{8}\\x=-\dfrac{9}{4}:2=-\dfrac{9}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{8};-\dfrac{9}{8}\right\}\)
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)
\(\dfrac{8}{5}\).\(\dfrac{6}{19}\) + \(\dfrac{13}{19}\).\(\dfrac{16}{19}\) + \(\dfrac{2}{5}\)
= \(\dfrac{6}{19}\).(\(\dfrac{8}{5}\) + \(\dfrac{13}{19}\)) + \(\dfrac{2}{5}\)
= \(\dfrac{6}{19}\).\(\dfrac{217}{95}\) + \(\dfrac{2}{5}\)
= \(\dfrac{19520}{1805}\) + \(\dfrac{2}{5}\)
= \(\dfrac{2674}{1805}\)