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\(\Rightarrow B=-1.-1^3.....-1^{2013}\left(-1^{2x}=1\right).\)
\(=-1^{1008}\)
= 1
\(\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^3:\left(\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{\left(\frac{2}{5}.5\right)^7+\left(\frac{9}{4}:\frac{3}{16}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+\left(\frac{9}{4}.\frac{16}{3}\right)^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.4^3}{2^7.5^2.2^9}\)
=\(\frac{2^7+3^3.2^6}{2^7.5^2.2^9}\)
=\(\frac{2^6.\left(27+2\right)}{2^6.5^2.2^{10}}\)
=\(\frac{29}{25600}\)
Đoạn \(\frac{2^6\cdot\left(27+2\right)}{2^6\cdot5^2\cdot2^{10}}\)là sai rồi bn ơi!!!
Bn phải lm như mục trên là:
\(\frac{2^6\left(2+3^3\right)}{2^7\left(5^2+2^2\right)}\)\(=\frac{2^6\cdot29}{2^7\cdot29}=\frac{1}{2}\)
Nhưng dù sao cx c.ơn bn vì đã giúp mk,mk sẽ cho bn 1 ths nka!!!
a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)
\(=4\cdot5-\dfrac{1+8}{3}\)
=20-3
=17
b: \(5^8:5^6+4\left(3^2-1\right)\)
\(=5^2+4\left(9-1\right)\)
=25+4*8
=25+32
=57
c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)
\(=400-36+20:\left[27-5\right]\)
\(=364+\dfrac{20}{22}\)
\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)
A) 2².5 - (1¹⁰ + 8) : 3²
= 4.5 - (1 + 8) : 9
= 20 - 9 : 9
= 20 - 1
= 19
B) 5⁸ : 5⁶ + 4.(3² - 1)
= 5² + 4.(9 - 1)
= 25 + 4.8
= 25 + 32
= 57
C) 400 - {36 - 20 : [3³ - (8 - 3)]}
= 400 - [36 - 20 : (27 - 5)]
= 400 -(36 - 20 : 22)
= 400 - (36 - 10/11)
= 400 - 386/11
= 4014/11
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
a) 2011 + 5[300 - (17 - 7)2 ]
= 2011 + 5[300 - 100]
= 2011 + 5.200
= 2011 + 1000
= 3011
b) 695 - [200 + (11 - 1)2 ]
= 695 - [200 + 100]
= 695 - 300
= 395
c) 129 - 5[29 - (6-1)2 ]
= 129 - 5[29 - 25]
= 129 - 5 . 4
= 129 - 20
= 109
d) 2345 - 1000 : [19 - 2(21-18)2 ]
= 2345 - 1000 : [19 - 2 . 9]
= 2345 - 1000 : 1
= 2345 - 1000
= 1345
a, 2011+5 [300-(17-7)2 ] =2011+5[300-102]
=2011+5[300-100]=2011+5.200
=2011+1000=3011
b, 695-[200+(11-1)2]=695-[200+102]
=695-[200+100]=695-300=395
