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\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)
(Ko chép lại đề nữa nhé, đánh đề bài xoắn cả tay)
\(P=\dfrac{\left(17,005-4,505\right)^2+125,075}{\left\{\left[0,1936:0,88+3,53\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{\left(12,5\right)^2+125,075}{\left\{\left[3,75\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{156,25+125,075}{\left\{14,0625-7,5625\right\}:0,52}\)
\(=\dfrac{281,325}{6,5:0,52}\)
\(=\dfrac{281,325}{12,5}\)
\(=22,506\)
\(Q=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\left(\dfrac{1431}{108}-\dfrac{236}{108}-\dfrac{1170}{108}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{30}{21}+\dfrac{70}{21}\right):\left(\dfrac{259}{21}-\dfrac{300}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{1016}{5}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-1}{21}}\)
\(=\dfrac{\dfrac{1270}{27}+\dfrac{187}{4}}{-100}\)
\(=\dfrac{\dfrac{5080}{108}+\dfrac{5049}{108}}{-100}\)
\(=\dfrac{10129}{108}.\left(-\dfrac{1}{100}\right)\)
\(=-\dfrac{10129}{10800}\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)\(=\dfrac{\dfrac{2^3\cdot\left(-3\right)^2\cdot\left(-1\right)}{3^3\cdot4^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot12^3}}\)
\(=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot\left(2^2\right)^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot\left(2^2\cdot3\right)^3}}=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot2^4}}{\dfrac{2^2\cdot5^2\cdot\left(-5\right)}{5^2\cdot2^6\cdot3^3}}=\dfrac{\dfrac{-1}{3\cdot2}}{\dfrac{-5}{2^4\cdot3^3}}\)
\(=\dfrac{-\dfrac{1}{6}}{27\cdot16}=-\dfrac{1}{6}:432=-\dfrac{1}{6}\cdot\dfrac{1}{432}=-\dfrac{1}{2592}\)
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^2}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\left(\dfrac{2}{5}.-\dfrac{5}{12}\right)^2}\)
\(=\dfrac{\dfrac{1}{3}.\dfrac{1}{2}.\left(-1\right)}{\left(-\dfrac{1}{6}\right)^2}=\dfrac{-\dfrac{1}{6}}{\dfrac{1}{36}}=-6\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^{2011}}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)
\(B=\dfrac{\dfrac{2}{3}\cdot\left(-\dfrac{3}{4}\cdot\dfrac{2}{3}\right)^2\cdot\left(-1\right)}{-\dfrac{5}{12}\left(-\dfrac{5}{12}\cdot\dfrac{2}{5}\right)^2}\)
\(B=\dfrac{-\dfrac{2}{3}\cdot\dfrac{1}{4}}{-\dfrac{5}{12}\cdot\dfrac{1}{36}}=-\dfrac{1}{6}:-\dfrac{5}{432}\)
\(B=\dfrac{72}{5}\)