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Giải:
a) \(11\dfrac{3}{4}.\left(6\dfrac{5}{6}-4\dfrac{1}{2}+1\dfrac{2}{3}\right)\)
\(=\dfrac{47}{4}.\left(\dfrac{41}{6}-\dfrac{9}{2}+\dfrac{5}{3}\right)\)
\(=\dfrac{47}{4}.4\)
\(=47\)
b) \(\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)
\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\dfrac{25}{8}:\dfrac{75}{26}\)
\(=\dfrac{13}{12}\)
c) \(\left(17\dfrac{13}{15}-3\dfrac{3}{7}\right)-\left(2\dfrac{12}{15}-4\right)\)
\(=\dfrac{268}{15}-\dfrac{24}{7}-\dfrac{14}{5}+4\)
\(=\left(\dfrac{268}{15}-\dfrac{14}{5}\right)+\left(\dfrac{-24}{7}+4\right)\)
\(=\dfrac{226}{15}+\dfrac{4}{7}\)
\(=\dfrac{1642}{105}\)
d) \(2\dfrac{2}{3}.\left(\dfrac{-4}{5}.0,375.-10.\dfrac{-15}{24}\right)\)
\(=\dfrac{8}{3}.\left(\dfrac{-4}{5}.\dfrac{3}{8}.-10.\dfrac{-5}{8}\right)\)
\(=\left(\dfrac{8}{3}.\dfrac{3}{8}\right).\left(\dfrac{-4}{5}.\dfrac{-5}{8}.-10\right)\)
\(=1.-5\)
\(=-5\)
Chúc bạn học tốt!
1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3
2:
1: =>36x+14x=69+81=150
=>50x=150
=>x=3
2: 3^x=81
=>3^x=3^4
=>x=4
3: 3(2x+1)^2=75
=>(2x+1)^2=25
=>2x+1=5 hoặc 2x+1=-5
=>x=-3 hoặc x=2
1:
1: \(\dfrac{13\cdot17^4+4\cdot17^4}{17^3}-\dfrac{14\cdot3^3-14\cdot3^2}{9}\)
\(=\dfrac{17^4\cdot\left(13+4\right)}{17^3}-\dfrac{14\cdot3^2\left(3-1\right)}{9}\)
\(=17\cdot17-14\cdot2\)
=289-28
=261
2:
\(2^3\cdot5^2-\left[131-\left(23-2^3\right)^2\right]\)
\(=8\cdot25-131+\left(-1\right)^2\)
=69+1
=70
\(\left(1\right)\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=-\dfrac{7}{12}.\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=-\dfrac{7}{12}.6+\dfrac{1}{2}=-3.\)
\(\left(2\right)\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4-\dfrac{2}{5}=\dfrac{-5}{3}-\dfrac{2}{5}=\dfrac{-31}{15}.\)
\(\left(3\right)\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=1-1-\dfrac{3}{2}=-\dfrac{3}{2}.\)
1. \(\dfrac{-7}{12}.\dfrac{11}{8}-\dfrac{37}{8}.\dfrac{7}{12}+\dfrac{1}{2}=\dfrac{-7}{12}\left(\dfrac{11}{8}+\dfrac{37}{8}\right)+\dfrac{1}{2}=\dfrac{-7}{12}.\dfrac{6}{1}+\dfrac{1}{2}=\dfrac{-7}{2}+\dfrac{1}{2}=\dfrac{-6}{2}=-3\)2.
\(\left(\dfrac{2}{3}-\dfrac{1}{4}-\dfrac{5}{6}\right).\left(-2\right)^2+\dfrac{3}{2}:\dfrac{-15}{4}=\dfrac{-5}{12}.4+\dfrac{-2}{5}=\dfrac{-5}{3}+\dfrac{-2}{5}=\dfrac{-31}{15}\)
3.
\(\dfrac{-2}{5}+\dfrac{3}{10}-\dfrac{3}{5}+\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{-4}{10}+\dfrac{3}{10}-\dfrac{6}{10}+\dfrac{7}{10}-\dfrac{15}{10}=\dfrac{-15}{10}=\dfrac{-3}{2}\)