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a. \(5^{10}.125^2.625^3=5^{10}.\left(5^3\right)^2.\left(5^4\right)^3=5^{10}.5^6.5^{12}=5^{10+6+12}=5^{28}\)
a: \(=5^{10}\cdot5^6\cdot5^{12}=5^{28}\)
b: \(=10^3\cdot10^8\cdot10^{15}=10^{26}\)
c: \(=2^{20}\cdot2^{20}=2^{40}\)
d: \(=2^{16}\cdot2^{16}\cdot3^8=2^{32}\cdot3^8\)
e: \(=\dfrac{3^{24}}{3^8}=3^{16}\)
f: \(=2^{12}\cdot2^{20}\cdot2^5=2^{37}\)
a/=(74-(-1937)1)
=74-(-1937)
=2011
b/=4/7+5/6:5-3/8*(-4)
=4/7+1/6-(-3/2)
=31/42-(-3/2)
=47/21
minh chi biet bay nhieu
Bài làm
a) \(-\frac{3}{7}+\frac{3}{4}:\frac{3}{14}\)
= \(-\frac{3}{7}+\frac{3}{4}.\frac{14}{3}\)
= \(-\frac{3}{7}+\frac{7}{2}\)
\(=-\frac{7}{14}+\frac{49}{14}\)
\(=\frac{42}{14}=3\)
b) \(5-\frac{7}{39}:\frac{7}{13}+\frac{8}{9}:4\)
\(=5=\frac{7}{39}.\frac{13}{7}+\frac{8}{9}.\frac{1}{4}\)
\(=5-\frac{1}{3}+\frac{2}{9}\)
\(=\frac{45}{9}-\frac{3}{9}+\frac{2}{9}\)
\(=\frac{44}{9}\)
c) \(\left(\frac{5}{12}:\frac{11}{6}+\frac{5}{12}:\frac{11}{5}\right)-\frac{-7}{12}\)
\(=\left(\frac{5}{12}.\frac{6}{11}+\frac{5}{12}.\frac{5}{11}\right)+\frac{7}{12}\)
\(=\left[\frac{5}{12}\left(\frac{6}{11}+\frac{5}{11}\right)\right]+\frac{7}{12}\)
\(=\frac{5}{12}+\frac{7}{12}\)
\(=\frac{12}{12}=1\)
d) \(-\frac{5}{9}+\frac{14}{9}\left(\frac{3}{4}-\frac{2}{5}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}\left(\frac{15}{20}-\frac{8}{20}\right):49\)
\(=-\frac{5}{9}+\frac{14}{9}.\frac{7}{20}.\frac{1}{49}\)
\(=-\frac{5}{9}+\frac{7}{9}.\frac{7}{10}.\frac{1}{7.7}\)
\(=-\frac{5}{9}+\frac{1}{90}\)
\(=-\frac{50}{90}+\frac{1}{90}=-\frac{49}{90}\)
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
a;\(\dfrac{-6}{11}\) : \(\dfrac{12}{55}\) = \(\dfrac{-5}{2}\)
b;\(\dfrac{7}{12}\) + \(\dfrac{5}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{47}{72}\) - \(\dfrac{11}{36}\) = \(\dfrac{25}{72}\)
c;\(\dfrac{13}{10}\) : \(\dfrac{-5}{13}\) = \(\dfrac{-169}{50}\)
d; {\(\dfrac{5}{12}\) + \(\dfrac{5}{11}\) } : { \(\dfrac{5}{3}\) -\(\dfrac{7}{11}\) } = \(\dfrac{115}{132}\) : \(\dfrac{34}{33}\) = \(\dfrac{115}{136}\)
lưu ý mk ko chép đầu bài
mình cần gấp lắm đến chiều mai là phải nộp rùi
giúp mình nha thanks cá bạn trước 
ko có tâm trạng mà cười nữa
Bài đây tính nhanh nhé ミ★ʟuғғʏ☆мũ☆ʀơм★彡 chứ không phải quy đồng lên đâu :)
a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)
\(A=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=35-\frac{167}{32}=\frac{953}{32}\)
b) \(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}:\frac{-7}{3}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot\frac{-3}{7}+2\frac{3}{7}\)
\(B=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)
\(B=\frac{-3}{7}+\frac{17}{7}=\frac{14}{7}=2\)
c) \(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{2}{8}\right):\frac{7}{12}\right]\cdot\frac{4}{5}\)
\(C=6\frac{3}{8}\cdot\frac{4}{5}=\frac{51}{8}\cdot\frac{4}{5}=\frac{51}{2}\cdot\frac{1}{5}=\frac{51}{10}\)
d) \(D=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)\cdot107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
b1: a, 612.(15+19-34)=612.0=0
b,414.(37.4+23.4-240)=414.0=0
c,(517.125-518.25)+63:23=(517.53-518.52)+33=0+27=27
b2:a,143+7.(n-17)=206
===> 7.(n-17)=206-143=63
====>n-17=63:7=9
=====>n=9+17=26
vậy n=26
b,128-28:(15-n)=124
====>28:(15-n)=128-124=4
=====> 15-n=28:4=7
=====> n=15-7=8
vậy n=8
c,3n.2+48=210
====>3n.2=210-48=162
====>3n=162:2=81=34
====>n=4
vậy n=4
a)75:73+32x29-20160
=72+32+29-20160
=49+9+(22+22+22+23)-1
=49+9+(4+4+4+8)-1
=49+9+20-1
=77
a) \(5^6.4^3.19=5^6.\left(2^2\right)^3.19=\left(5.2\right)^6.19=1000000.19=19000000\)
b)\(123.25^4.16^2=123.5^8.2^8=123.\left(5.2\right)^8=123.10^8=123.100000000=12300000000\)
c) \(73.5^{12}.8^4=73.5^{12}.\left(2^3\right)^4=73.\left(5.2\right)^{12}=73.10^{12}=73.1000000000000=73000000000000\)
d)\(1111.8^4.25^7=1111.\left(2^3\right)^4.\left(5^2\right)^7=1111.2^{12}.5^{14}=\left(1111.25\right).\left(2.5\right)^{12}=27775.10^{12}\)
\(=27775.1000000000000=27775000000000000\)
a)56x43=1000000
1000000x19=19000000
b) 12300000000
c) 73000000000000
d) Ko ra đc