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a: -8/31=-808/3131
-786/3131=-786/3131
b: \(\dfrac{11}{2^3\cdot3^4\cdot5^2}=\dfrac{11\cdot5}{2^3\cdot3^4\cdot5^3}=\dfrac{55}{2^3\cdot3^4\cdot5^3}\)
\(\dfrac{29}{2^2\cdot3^4\cdot5^3}=\dfrac{29\cdot2}{2^3\cdot3^4\cdot5^3}=\dfrac{58}{2^3\cdot3^4\cdot5^3}\)
c: 7/39=140/780
11/65=132/780
9/52=135/780
a: \(\dfrac{4\cdot5+4\cdot11}{8\cdot7+4\cdot3}=\dfrac{20+44}{56+12}=\dfrac{64}{68}=\dfrac{16}{17}=\dfrac{11088}{11781}\)
\(\dfrac{-15\cdot8+10\cdot7}{5\cdot6+20\cdot3}=\dfrac{-120+70}{30+60}=\dfrac{-50}{90}=\dfrac{-5}{9}=\dfrac{-6545}{11781}\)
\(\dfrac{2^4\cdot5^2\cdot7}{2^3\cdot5\cdot7^2\cdot11}=\dfrac{2\cdot5}{7\cdot11}=\dfrac{10}{77}=\dfrac{1530}{11781}\)
Câu 1:
\(\dfrac{14\cdot34-21\cdot10}{28\cdot10-28\cdot16}=\dfrac{7\cdot2\cdot34-7\cdot3\cdot10}{28\left(10-16\right)}\)
\(=\dfrac{7\left(2\cdot34-3\cdot10\right)}{28\cdot\left(-6\right)}=\dfrac{1}{4}\cdot\dfrac{68-30}{-6}=\dfrac{1}{4}\cdot\dfrac{38}{-6}=\dfrac{38}{-24}=-\dfrac{19}{12}\)
a: \(\dfrac{25\cdot9-25\cdot17}{-8\cdot80-8\cdot10}=\dfrac{25\cdot\left(-8\right)}{\left(-8\right)\cdot\left(80+10\right)}=\dfrac{25}{90}=\dfrac{5}{18}=\dfrac{250}{900}\)
\(\dfrac{48\cdot12-48\cdot15}{\left(-3\right)\cdot\left(270+30\right)}=\dfrac{48}{300}=\dfrac{8}{50}=\dfrac{144}{900}\)
\(\dfrac{2^5\cdot3}{600}=\dfrac{96}{600}=\dfrac{144}{900}\)
b: \(\dfrac{2^5\cdot7+25}{25\cdot5^2-2^5\cdot3}=\dfrac{32\cdot7+25}{25\cdot25-32\cdot3}=\dfrac{249}{529}=\dfrac{1743}{7\cdot529}\)
\(\dfrac{3^4\cdot5-3^6}{3^4\cdot13+3^4}=\dfrac{3^4\left(5-9\right)}{3^4\left(13+1\right)}=\dfrac{-4}{14}=\dfrac{-2}{7}=\dfrac{-1058}{7\cdot529}\)
1,
B=\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.........+\(\dfrac{1}{2^{2017}}\)
2B=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\)
2B-B=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+.......+\(\dfrac{1}{2^{2016}}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+.......+\(\dfrac{1}{2^{2017}}\))
B=1-\(\dfrac{1}{2^{2017}}\)
Vậy B=1-\(\dfrac{1}{2^{2017}}\)
Ta có: \(\dfrac{4.5^2}{5^3.6}=\dfrac{2.2.5^2}{5^2.5.2.3}=\dfrac{2}{15}\)
\(\dfrac{15}{1.2.3.4.5}=\dfrac{3.5}{1.2.3.4.5}=\dfrac{1}{8}\)
Quy đồng: \(\dfrac{2}{15}=\dfrac{16}{120}\) ; \(\dfrac{1}{8}=\dfrac{15}{120}\)
Vậy \(\dfrac{4.5^2}{5^3.6}=\dfrac{16}{120}\) và \(\dfrac{15}{1.2.3.4.5}=\dfrac{15}{120}\)
Ta có:\(\dfrac{x}{-12}=\dfrac{-3}{x}\)
\(\Rightarrow x.x=-3.\left(-12\right)\)
\(x^2=36\)
Vì \(x\in Z\)\(\Rightarrow x=\pm6\)
bài 4 dễ mà , bạn làm xong rồi gửi cho mik , đễ mik xem có đúng k nhé
Ghi rõ đề ra chứ :((
Tớ ghi rõ rồi mà