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B1
a) \(1-\left(5\frac{3}{8}+x-7\frac{5}{24}\right):16\frac{2}{3}=0\)
\(1-\left(\frac{43}{8}+x-\frac{173}{24}\right):\frac{50}{3}=0\)
\(1-\left(x-\frac{11}{6}\right).\frac{3}{50}=0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1-0\)
\(\left(x-\frac{11}{6}\right).\frac{3}{50}=1\)
\(x-\frac{11}{6}=1:\frac{3}{50}\)
\(x-\frac{11}{6}=\frac{50}{3}\)
\(x=\frac{50}{3}+\frac{11}{6}\)
\(x=\frac{37}{2}\)
b) \(\frac{3}{5}+\frac{5}{7}:x=\frac{1}{3}\)
\(\frac{5}{7}:x=\frac{1}{3}-\frac{3}{5}\)
\(\frac{5}{7}:x=-\frac{4}{15}\)
\(x=\frac{5}{7}:\left(-\frac{4}{15}\right)\)
\(x=-\frac{75}{28}\)
c) \(\left(4\frac{1}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\left(\frac{9}{2}-\frac{2}{5}.x\right):\frac{7}{4}=\frac{11}{9}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{9}.\frac{7}{4}\)
\(\frac{9}{2}-\frac{2}{5}.x=\frac{11}{2}\)
\(\frac{2}{5}.x=\frac{9}{2}-\frac{11}{2}\)
\(\frac{2}{5}.x=-1\)
\(x=-1:\frac{2}{5}\)
\(x=-\frac{5}{2}\)
B2
a) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{6}\right).24:5-\frac{9}{22}:\frac{15}{121}\)
\(=\left(\frac{3}{6}+\frac{2}{6}+\frac{2}{6}\right).24:5-\frac{9}{22}.\frac{121}{15}\)
\(=\frac{7}{6}.24:5-\frac{33}{10}\)
\(=28:5-\frac{33}{10}\)
\(=\frac{28}{5}-\frac{33}{10}\)
\(=\frac{56}{10}-\frac{33}{10}\)
\(=\frac{23}{10}\)
b) \(\frac{5}{14}+\frac{18}{35}+\left(1\frac{1}{4}-\frac{5}{4}\right):\left(\frac{5}{12}\right)^2\)
\(=\frac{25}{70}+\frac{36}{70}+\left(\frac{5}{4}-\frac{5}{4}\right):\frac{25}{144}\)
\(=\frac{61}{70}+0:\frac{25}{144}\)
\(=\frac{61}{70}+0\)
\(=\frac{61}{70}\)
a, A = \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)
\(A=\frac{2^{10}\left(13+65\right)}{2^8.2^2.26}=\frac{2^{10}.78}{2^{10}.26}=\frac{78}{26}=3\)
Vậy A = 3
b, \(B=\frac{72^3.54^2}{108^4}=\frac{72^3.54^2}{\left(54.2\right)^4}=\frac{72^3.54^2}{54^4.2^4}=\frac{72^3}{54^2.2^4}=\frac{\left(8.9\right)^3}{\left(6.9\right)^2.2^4}\)
\(=\frac{\left(2^3\right)^3.9^3}{6^2.9^2.2^4}=\frac{2^9.9^3}{2^2.3^2.9^2.2^4}=\frac{2^9.9^3}{2^6.9^3}=\frac{2^9}{2^6}=2^3=8\)
Vậy B = 8
c, \(C=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}.3^{30}}{2^2.3^{28}}=\frac{11.3^{29}.3.3^{29}}{2^2.3^{28}}=\frac{\left(11-3\right)3^{29}}{2^2.3^{28}}\)
\(=\frac{2^3.3^{29}}{2^2.3^{28}}=2.3=6\)
Vậy C = 6
d, \(D=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^{18}\right)^2}{11.2^{35}-\left(2^4\right)^9}=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{\left(11-2\right)2^{35}}=\frac{3^2.2}{9}=2\)
Vậy D = 2
A= 3^10.( 11+5 ) / 3^9. 2^4
A= 3^10. 16 /3 ^9 . 16
A= 3^10/3^9= 3
B= 2^10. ( 13 +65) / 2^8.104
B= 2^10. 78/ 2^8.104
B= 2 ^10.2.39/ 2^8 .2.52
B= 2^11.39/ 2^9.52
B= 2 ^ 2. (39/ 52)
B= 4 . 39/52 = 3
C= (2^3.3^2)^3.( 2.3.3^2 )^2 / (2^2.3^3)^4
C= 2^9.3^6/ 2^2.3^2.3^4
C= 2^9.3^6/ 2^2.3^6
C= 2^9/2^2= 2^5=32
D= 11.3^29-3^30/ 2^2.3^28
D= 3^29.(1- 3)/ 2^2.3^28
D= 3^29.(-2)/ 2^2.3^28
D= 3. (-2/2^2)
D = 3. (-1/2)= -3/2
S = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + (9 - 10 - 11 + 12) + (13 - 14 - 15 + 16) + (17 - 18)
= 0 + 0 + 0 + 0 + (-1)
= -1
A = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (2013 - 2014 - 2015 + 2016) + 2017
= 0 + 0 + ... + 0 + 2017
= 2017
S = (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) +...... + (13 - 14 - 15 + 16)+ 17 - 18
S = 17 - 18 = -2
A = (1 - 2 - 3 + 4) + (5 - 6 -7 + 8) + ..... + (2013 - 2014 - 2015 + 2016) + 2017
A = 2017
a) 378
b) 3
c) 2
d) 2
e) \(\frac{8748}{1715}\)
Mình thấy bài e) bạn có ghi thiếu ko vậy.81^2 x;: hay là cộng trừ vậy?
a) 11 22 − 3 16 . 8 18
= 1 2 − 1 12 = 5 12
b) 2 3 + 4 5 . 10 4
= 2 3 + 2 = 2 2 3
c) 1 3 + 4 6 . 2 7 + 9 14
= 1 3 + 4 6 . 2 7 + 9 14 = 1 3 + 2 3 . 4 14 + 9 14 = 1. 13 14 = 13 14