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4.{\(^{3^2}\).[(\(^{5^2}\)+\(^{2^3}\)):11]-26}+2009=...
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C
10 tháng 12 2023
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
17 tháng 10 2021
1.a) 4.(12+88) = 4.100=400
b)72- 6 .[2.8:8 ]=72-6.[16:8]=72-6 .2= 49-12=37 ( tui ko chắc lắm )
2.a) x=26-11=15
b) (4x-15)3=53
4x-15=5
4x= 5+15
4x=20
vậy x = 20
17 tháng 10 2021
Bài 2:
a: Ta có: x+11=26
nên x=15
b: Ta có: \(\left(4x-15\right)^3=125\)
\(\Leftrightarrow4x-15=5\)
hay x=5
18 tháng 4 2023
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
3 tháng 7 2021
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!
L
4
14 tháng 2 2016
1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 +…+ 2009 – 2010 – 2011 + 2012.
= (1 – 2 – 3 + 4) +( 5 – 6 – 7 + 8) +…+ (2009 – 2010 – 2011 + 2012)
= 0 + 0 + ...+ 0 = 0
LS
0
NT
1
S
1
5 tháng 11 2021
\(=\dfrac{11}{17}:\dfrac{19}{40}+\dfrac{11}{17}:\dfrac{23}{20}=\dfrac{11}{17}\cdot\left(\dfrac{40}{19}+\dfrac{20}{23}\right)=\dfrac{11}{17}\cdot\dfrac{1300}{437}=\dfrac{14300}{7429}\)
KT
27 tháng 3 2018
\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}-1-\frac{7}{11}}\)
\(=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}-\frac{132}{132}+\frac{84}{132}}\)
\(=\frac{\frac{115}{132}}{\frac{7}{132}}\)
\(=\frac{115}{132}\div\frac{7}{132}\)
\(=\frac{115}{7}\)
\(4.\left\{3^2.\left[\left(5^2+2^3\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[\left(25^{ }+8\right):11\right]-26\right\}+2009\)
\(=4.\left\{9.\left[33:11\right]-26\right\}+2009\)
\(=4.\left\{9.3-26\right\}+2009\) \(=4.\left(27-26\right)+2009\)
\(=4.1+2009=2013\)