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\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)
\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
a) \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(\dfrac{-4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b) \(B=2\dfrac{3}{11}.1\dfrac{1}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{13}{12}.\dfrac{-11}{5}=-\dfrac{65}{12}\)
c) \(C=\left(\dfrac{3}{4}-0,2\right)\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right)\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}\left(\dfrac{-2}{5}\right)=\dfrac{-11}{50}\)
a) \(A=\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right):\left(-3\right)\)
\(A=\left(-0,75-0,25\right):\left(-5\right)+\dfrac{1}{48}-\left(-\dfrac{1}{6}\right)\cdot\dfrac{-1}{3}\)
\(A=\left(-1\right):\left(-5\right)+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{1}{5}+\dfrac{1}{48}-\dfrac{1}{18}\)
\(A=\dfrac{119}{720}\)
b) \(B=\left(\dfrac{6}{25}-1,24\right):\dfrac{3}{7}:\left[\left(3\dfrac{1}{2}-3\dfrac{2}{3}\right):\dfrac{1}{14}\right]\)
\(B=\left(0,24-1,24\right):\dfrac{3}{7}:\left[\left(\dfrac{7}{2}-\dfrac{11}{3}\right):\dfrac{1}{14}\right]\)
\(B=-1:\dfrac{3}{7}:\left(-\dfrac{1}{6}:\dfrac{1}{14}\right)\)
\(B=-\dfrac{7}{3}:-\dfrac{7}{3}\)
\(B=1\)
a, A = (-0,75 - \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{48}\) - (- \(\dfrac{1}{6}\)) : (-3)
A = -(0,75 + 0,25): (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = -1 : (-5) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{1}{5}\) + \(\dfrac{1}{48}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{53}{240}\) - \(\dfrac{1}{18}\)
A = \(\dfrac{119}{720}\)
b, B = (\(\dfrac{6}{25}\) - 1,24): \(\dfrac{3}{7}\): [(3\(\dfrac{1}{2}\) - 3\(\dfrac{2}{3}\)): \(\dfrac{1}{14}\)]
B = (0,24 - 1,24): \(\dfrac{3}{7}\):[(\(\dfrac{7}{2}\)-\(\dfrac{11}{3}\)): \(\dfrac{1}{14}\)]
B = -1: \(\dfrac{3}{7}\):[ (-\(\dfrac{1}{6}\) : \(\dfrac{1}{14}\))]
B = -1: \(\dfrac{3}{7}\): (- \(\dfrac{7}{3}\))
B = 1 \(\times\) \(\dfrac{7}{3}\) \(\times\) \(\dfrac{3}{7}\)
B = 1
\(\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)\(=\dfrac{\dfrac{2^3\cdot\left(-3\right)^2\cdot\left(-1\right)}{3^3\cdot4^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot12^3}}\)
\(=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot\left(2^2\right)^2}}{\dfrac{2^2\cdot\left(-5\right)^3}{5^2\cdot\left(2^2\cdot3\right)^3}}=\dfrac{\dfrac{2^3\cdot3^2\cdot\left(-1\right)}{3^3\cdot2^4}}{\dfrac{2^2\cdot5^2\cdot\left(-5\right)}{5^2\cdot2^6\cdot3^3}}=\dfrac{\dfrac{-1}{3\cdot2}}{\dfrac{-5}{2^4\cdot3^3}}\)
\(=\dfrac{-\dfrac{1}{6}}{27\cdot16}=-\dfrac{1}{6}:432=-\dfrac{1}{6}\cdot\dfrac{1}{432}=-\dfrac{1}{2592}\)
(Ko chép lại đề nữa nhé, đánh đề bài xoắn cả tay)
\(P=\dfrac{\left(17,005-4,505\right)^2+125,075}{\left\{\left[0,1936:0,88+3,53\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{\left(12,5\right)^2+125,075}{\left\{\left[3,75\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{156,25+125,075}{\left\{14,0625-7,5625\right\}:0,52}\)
\(=\dfrac{281,325}{6,5:0,52}\)
\(=\dfrac{281,325}{12,5}\)
\(=22,506\)
\(Q=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\left(\dfrac{1431}{108}-\dfrac{236}{108}-\dfrac{1170}{108}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{30}{21}+\dfrac{70}{21}\right):\left(\dfrac{259}{21}-\dfrac{300}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{1016}{5}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-1}{21}}\)
\(=\dfrac{\dfrac{1270}{27}+\dfrac{187}{4}}{-100}\)
\(=\dfrac{\dfrac{5080}{108}+\dfrac{5049}{108}}{-100}\)
\(=\dfrac{10129}{108}.\left(-\dfrac{1}{100}\right)\)
\(=-\dfrac{10129}{10800}\)
Lời giải:
1.
$(\frac{5}{6})^{10}.(\frac{3}{10})^{10}=(\frac{5}{6}.\frac{3}{10})^{10}=(\frac{1}{4})^{10}$
$=\frac{1}{4^{10}}$
2.
$(\frac{4}{7})^{19}: (\frac{-12}{35})^{19}=(\frac{4}{7}: \frac{-12}{35})^{19}=(\frac{-5}{3})^{19}$
3.
$(\frac{-3}{7})^7:\frac{-3}{5}=\frac{(-3)^7}{7^7}.\frac{5}{-3}=\frac{5.(-3)^6}{7^7}=\frac{5.3^6}{7^7}$
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3\cdot\left(-\dfrac{3}{4}\right)^2\cdot\left(-1\right)^{2011}}{\left(\dfrac{2}{5}\right)^2\cdot\left(-\dfrac{5}{12}\right)^3}\)
\(B=\dfrac{\dfrac{2}{3}\cdot\left(-\dfrac{3}{4}\cdot\dfrac{2}{3}\right)^2\cdot\left(-1\right)}{-\dfrac{5}{12}\left(-\dfrac{5}{12}\cdot\dfrac{2}{5}\right)^2}\)
\(B=\dfrac{-\dfrac{2}{3}\cdot\dfrac{1}{4}}{-\dfrac{5}{12}\cdot\dfrac{1}{36}}=-\dfrac{1}{6}:-\dfrac{5}{432}\)
\(B=\dfrac{72}{5}\)