Tính:
A/ √(27-12√5)
B/ √(13+4√10)
C/ √8-3√32+√72
D/ 6√1√20-2√27+√125
E/ √(4-√7)-√(4-7)+√2
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a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
a: Ta có: \(\left(8\cdot5^7+5^6-5^5\right):5^5\)
\(=8\cdot5^2+5-1\)
\(=200+4=204\)
b: Ta có: \(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=3^{60}:3^{57}-3^{57}:3^{57}+5^{27}:5^{24}-5^{24}:5^{24}\)
\(=27-1+125-1\)
=150
a. (8,57 - 55 + 56) : 55
= (8,57 : 55) - (55 : 55) + (56 : 55)
= 1,72 - 1 + 5
= 2,89 - 1 + 5
= 6,89
b. (930 - 2719) : 357 + (1259 - 2512) : 524
= (930 : 357) - (2719 : 357) + (1259 : 524) - (2512 : 524)
= 33 - 1 + 125 - 1
= 27 - 1 + 125 - 1
= 150
c. (1012 + 511 . 29 - 513 - 28) : 4 . 55 . 106
= (1012 + 2,5 , 1010 - 513 - 28) : 1,25 . 1010
= (1012 : 1,25 . 1010) + (2,5 . 1010 : 1,25 . 1010) - (513 : 1,25 . 1010) - (28 : 1,25 . 1010)
= 80 + 2 - \(\dfrac{25}{256}\) - \(\dfrac{1}{48828125}\)
= 81,90234373 \(\approx\) 82